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http://math.stackexchange.com/questions/4238/finite-non-abelian-group-with-centre-but-no-outer-automorphism?answertab=active
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# Finite non-abelian group with centre but no outer automorphism
The two element group has a centre (of course it does, it's abelian) but no outer automorphisms. I can come up with a (convoluted) countably infinite non-abelian example. Is there a finite one? If so what is the smallest?
Is there a name for this sort of group? ("Outerless?" Since a group without a centre is "centreless", or "complete" if it has no outermorphisms either.)
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## 2 Answers
The smallest example is 2×AΓL(1,8) of order 336. A fairly standard permutation representation is generated by (+,-), (∞,0)(1,3)(2,6)(4,5), (1,2,3,4,5,6,7), and (1,2,4)(3,6,5), or you can use more GAP friendly points to get (9,10), (8,7)(1,3)(2,6)(4,5), (1,2,3,4,5,6,7), and (1,2,4)(3,6,5). AΓL(1,8) is a complete, solvable group with no normal subgroup of index 2.
I just checked the groups ordered by size, skipping nilpotent groups since they always have outer automorphisms. Probably one could be a bit smarter, looking for complete groups with no 2-quotients, but I wasn't sure the smallest example would be of the form 2×Complete.
There are examples like 2.Sz(8) which have no outer automorphisms, have a non-trivial center, but where the quotient by the center is not complete (so it is definitely not 2×Complete).
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Brilliant. Thanks! – anon Sep 9 '10 at 22:14
$C_2\times M_{11}$.
Steve
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Oh, good. Yes, that works. Is it the smallest? – anon Sep 8 '10 at 23:30
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http://cms.math.ca/10.4153/CJM-2011-049-0
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Canadian Mathematical Society
www.cms.math.ca
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# Lifting Quasianalytic Mappings over Invariants
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[PDF: 307KB]
http://dx.doi.org/10.4153/CJM-2011-049-0
Canad. J. Math. 64(2012), 409-428
Published:2011-07-15
Printed: Apr 2012
• Armin Rainer,
Fakultät für Mathematik, Universität Wien, A-1090 Wien, Austria
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Format: LaTeX MathJax PDF
## Abstract
Let $\rho \colon G \to \operatorname{GL}(V)$ be a rational finite dimensional complex representation of a reductive linear algebraic group $G$, and let $\sigma_1,\dots,\sigma_n$ be a system of generators of the algebra of invariant polynomials $\mathbb C[V]^G$. We study the problem of lifting mappings $f\colon \mathbb R^q \supseteq U \to \sigma(V) \subseteq \mathbb C^n$ over the mapping of invariants $\sigma=(\sigma_1,\dots,\sigma_n) \colon V \to \sigma(V)$. Note that $\sigma(V)$ can be identified with the categorical quotient $V /\!\!/ G$ and its points correspond bijectively to the closed orbits in $V$. We prove that if $f$ belongs to a quasianalytic subclass $\mathcal C \subseteq C^\infty$ satisfying some mild closedness properties that guarantee resolution of singularities in $\mathcal C$, e.g., the real analytic class, then $f$ admits a lift of the same class $\mathcal C$ after desingularization by local blow-ups and local power substitutions. As a consequence we show that $f$ itself allows for a lift that belongs to $\operatorname{SBV}_{\operatorname{loc}}$, i.e., special functions of bounded variation. If $\rho$ is a real representation of a compact Lie group, we obtain stronger versions.
Keywords: lifting over invariants, reductive group representation, quasianalytic mappings, desingularization, bounded variation
MSC Classifications: 14L24 - Geometric invariant theory [See also 13A50] 14L30 - Group actions on varieties or schemes (quotients) [See also 13A50, 14L24, 14M17] 20G20 - Linear algebraic groups over the reals, the complexes, the quaternions 22E45 - Representations of Lie and linear algebraic groups over real fields: analytic methods {For the purely algebraic theory, see 20G05}
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http://mathoverflow.net/questions/57426/are-there-any-non-linear-solutions-of-cauchys-equation-fxyfxfy-with
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## Are there any non-linear solutions of Cauchy’s equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be s.t. $f(x+y) = f(x) + f(y), \ \forall x, y$
It is quite obvious that this implies $f(cx)=cx$ for all $c \in \mathbb{Z}$ and even further: $\forall c \in \mathbb{Q}$
It is also known that using an infinite dimensional basis for $\mathbb{R}$ over $\mathbb{Q}$, it is possible using the Axiom of Choice to construct a function which is not linear.
My question is whether there exists a known solution not invoking the Axiom of Choice?
Edit: My questions are:
1. Is it not possible to find an explicit expression for a non-linear solution? Why?
2. Can existence of non-linear solutions be proven from ZF alone?
Reference: Cauchy's functional equation
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8
All Lebesgue measurable solutions are linear, so there are no nonlinear solutions if you assume the Axiom of Determinacy (en.wikipedia.org/wiki/Axiom_of_determinacy). – George Lowther Mar 5 2011 at 1:53
Ignas- This question isn't well-defined; are you asking if you can prove the existence of such solutions just ZF? (George shows the answer is no). If you want to edit the question to be clearer, and try posting on meta, you might convince people to reopen it. – Ben Webster♦ Mar 5 2011 at 4:03
Thank you for the comments. Yes, I probably should have been more clear, I actually wanted to ask whether: 1) it is possible to find an explicit expression for such a non-linear function (obvious answer: no, but I was interested in how to prove it isn't possible); 2) existence of solutions in ZF. – Ignas Mar 5 2011 at 18:58
1
@Ignas: Your edited version is interesting! I have voted to re-open. – Andres Caicedo Mar 5 2011 at 22:34
## 1 Answer
Ignas:
It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable.
(This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here.)
The point is that it is consistent that every set of reals is measurable, and therefore every function $f:{\mathbb R}\to{\mathbb R}$ is measurable.
The first to realize that it is possible using choice to construct a non-linear additive function was Hamel in 1905 ("Eine Basis aller Zahlen und die unstetigen Losungen der Functionalgleichung: $f(x+y)=f(x)+f(y)$", Math Ann 60 459-462); indeed, a Hamel basis of ${\mathbb R}$ over ${\mathbb Q}$ allows us to provide examples. Note that with a Hamel basis it is straightforward to build Vitali's standard example of a non-measurable set.
Solovay ("A model of set-theory in which every set of reals is Lebesgue measurable", Annals of Mathematics. Second Series (1970) 92 1–56) proved that, relative to the consistency of an inaccessible cardinal, via forcing we can prove that it is consistent with ZF + DC that all sets of reals are Lebesgue measurable (DC is the axiom of dependent choice, a weak version of the axiom of choice useful to develop the basic theory of analysis). It follows that, in Solovay's model, all additive functions are linear and therefore no "explicit" example is possible.
The other standard approach to models where all sets of reals are measurable is via determinacy. However, this requires a stronger commitment in terms of consistency strength (using $\omega$ Woodin cardinals we can force an inner model of determinacy). The two approaches are closely related: Assuming enough large cardinals, the inner model $L({\mathbb R})$ consisting of all sets constructible from reals is a model of determinacy, and a Solovay model. Note that this is a theorem (from enough large cardinals) rather than a consistency result, i.e., $L({\mathbb R})$ is a model of these statements, without needing to pass to a forcing extension.
Solovay's result does require an inaccessible cardinal, this is a result of Shelah ("Can you take Solovay's inaccessible away?", Israel Journal of Mathematics (1984) 48 1–47). This leaves the question of whether (the consistency of) ZF suffices to produce a model where all additive maps are linear. But this is now easily solved: In the same paper, Shelah proved that it is relatively consistent with ZF that all sets of reals have the property of Baire. The standard proofs that additivity and measurability give linearity work with "Baire measurability" instead of Lebesgue measurability.
Let me close, however, by pointing out that, in appropriate models of set theory, we can "explicitly" build additive, non-linear maps. More precisely, in nice inner models, we have explicitly definable well-orderings of the reals, from which such maps can be built. For example, in Gödel's inner model L of constructible sets, there is an easily definable Hamel basis (see this MO question), from which we can easily define such a function. "Easily" is formalized in terms of the projective hierarchy. The Hamel basis we obtain is $\Pi^1_1$ and the function is $\Sigma^1_2$.
(By the way, the statement "there is a discontinuous additive function" is form 366 in Howard-Rubin "Consequences of the axiom of choice". The highly recommended companion website does not reveal any additional choice-like implications regarding this form. In particular, it only mentions Solovay's model for its negation, but not Shelah's.)
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(The existence of a Hamel basis is not exactly what's needed to construct the Vitali set.) – François G. Dorais♦ Mar 6 2011 at 12:16
@François: Thanks. I edited the line in question. – Andres Caicedo Mar 6 2011 at 16:26
wow, thank you for a long and detailed answer, now I feel it was much worth it to request a re-open! – Ignas Mar 6 2011 at 20:35
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http://math.stackexchange.com/questions/140254/schwarz-christoffel-complex-mapping
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# Schwarz-Christoffel Complex Mapping
Verify that the Schwarz-Christoffel mapping of $\mathbb H$ onto the infinite half strip described by $|\Re(z)|<\frac \pi 2$ and $\Im(z)>0$ is given by the arcsine function.
What does that mean and how do I do it?
Schwarz-Christoffel formula for the half-plane $\mathbb H$ to the polygon with exterior angles described by coefficients $\beta_k$ is $$f(z)=A_1\int_0^z \frac 1{(w-x_1)^{\beta_1}(w-x_2)^{\beta_2}\cdots(w-x_n)^{\beta_n}}\ dw+A_2,\quad (z \in \mathbb H).$$
I realize this isn't the best question, but I'm not even sure what to ask.
Edit: An attempt to add more specific questions:
1. The $x_n$ in the integral are supposed to be vertex points of the pre-image. If the pre-image is the upper half plane, how do I find vertex points?
2. How do I get the integral to map to the described half plane?
3. Where does the arcsine fit in? What does it mean for the mapping created by an integral to be "given by the arcsine function"?
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4
So, you're asking why you're supposed to be interested in the mapping $$\int_0^z \frac1{\sqrt{1-w^2}} \mathrm dw$$ I take it? – J. M. May 3 '12 at 5:22
No. I'm wondering how to verify that the mapping of $\mathbb H$ onto the given half strip is given by the arcsine function. – Jeff May 3 '12 at 6:28
Yes, and the integral I gave is supposed to be the result of applying Schwarz-Christoffel to the region you're studying... – J. M. May 3 '12 at 6:30
How did you get that? Note that I edited my question, too. – Jeff May 3 '12 at 8:20
1
An infinite half-strip is a triangle on the Riemann sphere (with a vertex located at the "point at infinity" $\hat{\infty}$). Perhaps a sequence of triangles that converges to the strip in the limit will help you? – anon May 3 '12 at 8:47
show 5 more comments
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http://mathhelpforum.com/advanced-math-topics/168556-problems-continuous-dynamical-systems.html
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# Thread:
1. ## Problems with Continuous Dynamical Systems
Hi there, I'm new to this but I was wondering if you could help me with my understanding of Dynamical Systems.
I have a continuous dynamical system given by the equations:
dx1/dt = x1(x2-x3)
dx2/dt = x2(x3-x1)
dx3/dt = x3(x1-x3)
Which I need to analyse. There are a few problems I have but the main ones are centred around orbits.
My question is why can an orbit starting in the region x1 >= 0, x2 >= 0 , x3 >= 0 never cross one of the lines x_i = 0?
Any help on the matter would be greatly appreciated I'm drawing a complete blank with this and could really do with some help.
2. Proceed as you would with a differential equation: For example in your first equation isolating $x_1$ we get $\ln(x_1)=\int (x_2-x_3)dt+C_1$. Applying this to the other two, adding them up, condensing the constants and using that the logarithm opens products we arrive at $\ln (x_1x_2x_3)=\int (x_2-x_3)dt+C$ or equivalently $x_1x_2x_3=e^{C\int (x_2-x_3)dt}$ and since the exponential is always positive, none of the $x_i$ can be zero or change sign because of continuity.
PS. I'm not too familiar with dynamical systems so check everything carefully.
3. Originally Posted by wattsup
dx1/dt = x1(x2-x3)
dx2/dt = x2(x3-x1)
dx3/dt = x3(x1-x3)
My question is why can an orbit starting in the region x1 >= 0, x2 >= 0 , x3 >= 0 never cross one of the lines x_i = 0?
An alternative. For the given system $x'=v(x)$ we have:
$\begin{Bmatrix}v(\alpha,0,0)=(0,0,0)\\ v(0,\beta,0)=(0,0,0) \\ v(0,0,\gamma)=(0,0,-\gamma^2)\end{matrix}$
This means that all points of the $x_1$ and $x_2$ axes are equilibrium points and the $x_3$ axis decomposes into three orbits. You can conclude.
Fernando Revilla
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http://programmingpraxis.com/2009/07/21/pollards-p-1-factorization-algorithm/?like=1&source=post_flair&_wpnonce=27cac980df
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Programming Praxis
A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer
Pollard’s P-1 Factorization Algorithm
July 21, 2009
Fermat’s little theorem states that for any prime number $p$, and any other number $a$, $a^{p-1} \equiv 1 \pmod{p}$. Rearranging terms, we have $a^{p-1} - 1 \equiv 0 \pmod{p}$, which means that $p$ divides the left-hand side of the equation, thus $p$ is a factor of the left-hand side.
John Pollard used this idea to develop a factoring method in 1974. His idea is to choose a very large number and see if it has any factors in common with $a^{p-1} - 1$, thus giving a factor of $p$. A systematic way to test many very large numbers is by taking large factorials, which have many small factors within them. Thus, Pollard’s $p-1$ factorization algorithm is this:
1) Choose $b$, which is known as the smoothness bound, and calculate the product of the primes to $b$ by $k = \mathrm{lcm}(1,2,3,\ldots,b)$.
2) Choose a random integer $a$ such that $1 < a < n$.
3) Calculate $g=\gcd(a,n)$. If $g$ is strictly greater than 1, then it is a non-trivial factor of $n$. Otherwise, continue to the next step.
4) Calculate $d = \gcd(a^k - 1, n)$. If $1 < d < n$, then $d$ is a non-trivial factor of $n$. If $d=1$, go to Step 1 and choose a larger $b$. If $d=n$, go back to Step 2 and choose another $a$.
In Step 4, you can quit with failure instead of returning to Step 1 if$b$ becomes too large, where “too large” is probably somewhere around $10^6$; in that case, you will need to continue with some other factoring algorithm.
Your task is to write a function that factors integers by Pollard’s $p-1$ method. What are the factors of $2^{98}-1$? When you are finished, you are welcome to read or run a suggested solution, or to post your solution or discuss the exercise in the comments below.
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5 Comments »
5 Responses to “Pollard’s P-1 Factorization Algorithm”
1. veer said
July 23, 2009 at 5:54 AM
I wrote a prolog implementation ,which works now and then :), but not
always,most of the time it says ‘out of local stack’.
So i saved the result, factors are [4432676798593, 4363953127297, 127, 43, 3].
I would love to see the prolog code for the above problem.
2. Francois Saint-Jacques said
August 13, 2009 at 3:33 PM
a must be coprime to p, not any other number.
3. September 16, 2011 at 9:00 AM
[...] have studied John Pollard’s p−1 algorithm for integer factorization on two previous occasions, giving first the basic single-stage algorithm and later adding a second stage. In [...]
4. September 24, 2011 at 10:56 PM
`all_factors` repeatedly applies a factorizing method `factor` to `n`:
```let all_factors factor n =
let open Big_int in
let rec go l n =
if le_big_int n unit_big_int then l else
if is_even_big_int n then go (two_big_int :: l) (shift_right_big_int n 1) else
if is_prime n then (n :: l) else
let f = factor n in
go (go l f) (div_big_int n f)
in List.sort compare_big_int (go [] n)
```
Besides `modexp_big_int` and `random_big_int`, I need to compute lcms, in this case optimized by knowing that the first factor is always small:
```let lcm_int_big_int p q =
let open Big_int in
let d = gcd p (int_of_big_int (mod_big_int q (big_int_of_int p))) in
mult_int_big_int (p / d) q
```
Finally, Pollard’s p-1:
```let pollard_pm1 =
let open Big_int in
let rec go b k n =
let a = add_int_big_int 2 (random_big_int (add_int_big_int (-2) n)) in
let g = gcd_big_int a n in
if gt_big_int g unit_big_int then g else
let d = gcd_big_int (pred_big_int (modexp_big_int a k n)) n in
if eq_big_int d n then go b k n else
if eq_big_int d unit_big_int then
let b = succ b in
if b > 1_000_000 then failwith "pollard_pm1" else
go b (lcm_int_big_int b k) n
else d
in
let b = 7 in
let k = List.fold_right lcm_int_big_int (range 2 b)
(List.fold_right mult_int_big_int (primes b) unit_big_int) in
all_factors (go b k)
```
This method seems to be much, much slower than Pollard’s Rho.
5. programmingpraxis said
September 25, 2011 at 12:08 AM
It shouldn’t be much, much slower than Pollard’s rho. The two methods are complementary. The rho method is generally used to find small factors, say from 4 to 12 digits, after trial division has found factors up to 3 digits. The p-1 method is generally used after the rho method has run for a while, in the hope that you get lucky and find a large factor, say 15 to 20 digits or sometimes more.
It’s been a while since I programmed in the ML family of languages, so I’ll ask some questions instead of making some statements.
1) I see only one bound b. If you look at the more recent p-1 exercise that uses the improved standard continuation, you’ll see a much faster version of the algorithm that uses two bounds b1 and b2.
2) In the pollard_pm1 function, the line let k = …, do you recompute the lcm at each b?
3) I don’t understand the List.fold_right in the line after that. You seem to be taking the product of all the primes less than b, but I’m not sure why.
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http://en.m.wikipedia.org/wiki/D-brane
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# D-brane
String theory
Superstring theory
Theory
Concepts
Related topics
Scientists
In string theory, D-branes are a class of extended objects upon which open strings can end with Dirichlet boundary conditions, after which they are named. D-branes were discovered by Dai, Leigh and Polchinski, and independently by Hořava in 1989. In 1995, Polchinski identified D-branes with black p-brane solutions of supergravity, a discovery that triggered the Second Superstring Revolution and led to both holographic and M-theory dualities.
D-branes are typically classified by their spatial dimension, which is indicated by a number written after the D. A D0-brane is a single point, a D1-brane is a line (sometimes called a "D-string"), a D2-brane is a plane, and a D25-brane fills the highest-dimensional space considered in bosonic string theory. There are also instantonic D(-1)-branes, which are localized in both space and time.
## Theoretical background
The equations of motion of string theory require that the endpoints of an open string (a string with endpoints) satisfy one of two types of boundary conditions: The Neumann boundary condition, corresponding to free endpoints moving through spacetime at the speed of light, or the Dirichlet boundary conditions, which pin the string endpoint. Each coordinate of the string must satisfy one or the other of these conditions. There can also exist strings with mixed boundary conditions, where the two endpoints satisfy NN, DD, ND and DN boundary conditions. If p spatial dimensions satisfy the Neumann boundary condition, then the string endpoint is confined to move within a p-dimensional hyperplane. This hyperplane provides one description of a Dp-brane.
Although rigid in the limit of zero coupling, the spectrum of open strings ending on a D-brane contains modes associated with its fluctuations, implying that D-branes are dynamical objects. When $N$ D-branes are nearly coincident, the spectrum of strings stretching between them becomes very rich. One set of modes produce a non-abelian gauge theory on the world-volume. Another set of modes is an $N \times N$ dimensional matrix for each transverse dimension of the brane. If these matrices commute, they may be diagonalized, and the eigenvalues define the position of the $N$ D-branes in space. More generally, the branes are described by non-commutative geometry, which allows exotic behavior such as the Myers effect, in which a collection of Dp-branes expand into a D(p+2)-brane.
Tachyon condensation is a central concept in this field. Ashoke Sen has argued that in Type IIB string theory, tachyon condensation allows (in the absence of Neveu-Schwarz 3-form flux) an arbitrary D-brane configuration to be obtained from a stack of D9 and anti D9-branes. Edward Witten has shown that such configurations will be classified by the K-theory of the spacetime. Tachyon condensation is still very poorly understood. This is due to the lack of an exact string field theory that would describe the off-shell evolution of the tachyon.
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## Braneworld cosmology
This has implications for physical cosmology. Because string theory implies that the Universe has more dimensions than we expect—26 for bosonic string theories and 10 for superstring theories—we have to find a reason why the extra dimensions are not apparent. One possibility would be that the visible Universe is in fact a very large D-brane extending over three spatial dimensions. Material objects, made of open strings, are bound to the D-brane, and cannot move "at right angles to reality" to explore the Universe outside the brane. This scenario is called a brane cosmology. The force of gravity is not due to open strings; the gravitons which carry gravitational forces are vibrational states of closed strings. Because closed strings do not have to be attached to D-branes, gravitational effects could depend upon the extra dimensions at right angles to the brane.
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## D-brane scattering
When two D-branes approach each other the interaction is captured by the one loop annulus amplitude of strings between the two branes. The scenario of two parallel branes approaching each other at a constant velocity can be mapped to the problem of two stationary branes that are rotated relative to each other by some angle. The annulus amplitude yields singularities that correspond to the on-shell production of open strings stretched between the two branes. This is true irrespective of the charge of the D-branes. At non-relativistic scattering velocities the open strings may be described by a low-energy effective action that contains two complex scalar fields that are coupled via a term $\phi^2\chi^2$. Thus, as the field $\phi$ (separation of the branes) changes, the mass of the field $\chi$ changes. This induces open string production and as a result the two scattering branes will be trapped.
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## Gauge theories
The arrangement of D-branes constricts the types of string states which can exist in a system. For example, if we have two parallel D2-branes, we can easily imagine strings stretching from brane 1 to brane 2 or vice versa. (In most theories, strings are oriented objects: each one carries an "arrow" defining a direction along its length.) The open strings permissible in this situation then fall into two categories, or "sectors": those originating on brane 1 and terminating on brane 2, and those originating on brane 2 and terminating on brane 1. Symbolically, we say we have the [1 2] and the [2 1] sectors. In addition, a string may begin and end on the same brane, giving [1 1] and [2 2] sectors. (The numbers inside the brackets are called Chan-Paton indices, but they are really just labels identifying the branes.) A string in either the [1 2] or the [2 1] sector has a minimum length: it cannot be shorter than the separation between the branes. All strings have some tension, against which one must pull to lengthen the object; this pull does work on the string, adding to its energy. Because string theories are by nature relativistic, adding energy to a string is equivalent to adding mass, by Einstein's relation E = mc2. Therefore, the separation between D-branes controls the minimum mass open strings may have.
Furthermore, affixing a string's endpoint to a brane influences the way the string can move and vibrate. Because particle states "emerge" from the string theory as the different vibrational states the string can experience, the arrangement of D-branes controls the types of particles present in the theory. The simplest case is the [1 1] sector for a Dp-brane, that is to say the strings which begin and end on any particular D-brane of p dimensions. Examining the consequences of the Nambu-Goto action (and applying the rules of quantum mechanics to quantize the string), one finds that among the spectrum of particles is one resembling the photon, the fundamental quantum of the electromagnetic field. The resemblance is precise: a p-dimensional version of the electromagnetic field, obeying a p-dimensional analogue of Maxwell's equations, exists on every Dp-brane.
In this sense, then, one can say that string theory "predicts" electromagnetism: D-branes are a necessary part of the theory if we permit open strings to exist, and all D-branes carry an electromagnetic field on their volume.
Other particle states originate from strings beginning and ending on the same D-brane. Some correspond to massless particles like the photon; also in this group are a set of massless scalar particles. If a Dp-brane is embedded in a spacetime of d spatial dimensions, the brane carries (in addition to its Maxwell field) a set of d - p massless scalars (particles which do not have polarizations like the photons making up light). Intriguingly, there are just as many massless scalars as there are directions perpendicular to the brane; the geometry of the brane arrangement is closely related to the quantum field theory of the particles existing on it. In fact, these massless scalars are Goldstone excitations of the brane, corresponding to the different ways the symmetry of empty space can be broken. Placing a D-brane in a universe breaks the symmetry among locations, because it defines a particular place, assigning a special meaning to a particular location along each of the d - p directions perpendicular to the brane.
The quantum version of Maxwell's electromagnetism is only one kind of gauge theory, a U(1) gauge theory where the gauge group is made of unitary matrices of order 1. D-branes can be used to generate gauge theories of higher order, in the following way:
Consider a group of N separate Dp-branes, arranged in parallel for simplicity. The branes are labeled 1,2,...,N for convenience. Open strings in this system exist in one of many sectors: the strings beginning and ending on some brane i give that brane a Maxwell field and some massless scalar fields on its volume. The strings stretching from brane i to another brane j have more intriguing properties. For starters, it is worthwhile to ask which sectors of strings can interact with one another. One straightforward mechanism for a string interaction is for two strings to join endpoints (or, conversely, for one string to "split down the middle" and make two "daughter" strings). Since endpoints are restricted to lie on D-branes, it is evident that a [1 2] string may interact with a [2 3] string, but not with a [3 4] or a [4 17] one. The masses of these strings will be influenced by the separation between the branes, as discussed above, so for simplicity's sake we can imagine the branes squeezed closer and closer together, until they lie atop one another. If we regard two overlapping branes as distinct objects, then we still have all the sectors we had before, but without the effects due to the brane separations.
The zero-mass states in the open-string particle spectrum for a system of N coincident D-branes yields a set of interacting quantum fields which is exactly a U(N) gauge theory. (The string theory does contain other interactions, but they are only detectable at very high energies.) Gauge theories were not invented starting with bosonic or fermionic strings; they originated from a different area of physics, and have become quite useful in their own right. If nothing else, the relation between D-brane geometry and gauge theory offers a useful pedagogical tool for explaining gauge interactions, even if string theory fails to be the "theory of everything".
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## Black holes
Another important use of D-branes has been in the study of black holes. Since the 1970s, scientists have debated the problem of black holes having entropy. Consider, as a thought experiment, dropping an amount of hot gas into a black hole. Since the gas cannot escape from the hole's gravitational pull, its entropy would seem to have vanished from the universe. In order to maintain the second law of thermodynamics, one must postulate that the black hole gained whatever entropy the infalling gas originally had. Attempting to apply quantum mechanics to the study of black holes, Stephen Hawking discovered that a hole should emit energy with the characteristic spectrum of thermal radiation. The characteristic temperature of this Hawking radiation is given by
$T_{\rm H} = \frac{\hbar c^3}{8\pi GM k_B} \;\quad(\approx {1.227 \times 10^{23}\; kg \over M}\; K)$,
where G is Newton's gravitational constant, M is the black hole's mass and kB is Boltzmann's constant.
Using this expression for the Hawking temperature, and assuming that a zero-mass black hole has zero entropy, one can use thermodynamic arguments to derive the "Bekenstein entropy":
$S_{\rm B} = \frac{k_B 4\pi G}{\hbar c} M^2.$
The Bekenstein entropy is proportional to the black hole mass squared; because the Schwarzschild radius is proportional to the mass, the Bekenstein entropy is proportional to the black hole's surface area. In fact,
$S_{\rm B} = \frac{A k_B}{4 l_{\rm P}^2},$
where $l_{\rm P}$ is the Planck length.
The concept of black hole entropy poses some interesting conundra. In an ordinary situation, a system has entropy when a large number of different "microstates" can satisfy the same macroscopic condition. For example, given a box full of gas, many different arrangements of the gas atoms can have the same total energy. However, a black hole was believed to be a featureless object (in John Wheeler's catchphrase, "Black holes have no hair"). What, then, are the "degrees of freedom" which can give rise to black hole entropy?
String theorists have constructed models in which a black hole is a very long (and hence very massive) string. This model gives rough agreement with the expected entropy of a Schwarzschild black hole, but an exact proof has yet to be found one way or the other. The chief difficulty is that it is relatively easy to count the degrees of freedom quantum strings possess if they do not interact with one another. This is analogous to the ideal gas studied in introductory thermodynamics: the easiest situation to model is when the gas atoms do not have interactions among themselves. Developing the kinetic theory of gases in the case where the gas atoms or molecules experience inter-particle forces (like the van der Waals force) is more difficult. However, a world without interactions is an uninteresting place: most significantly for the black hole problem, gravity is an interaction, and so if the "string coupling" is turned off, no black hole could ever arise. Therefore, calculating black hole entropy requires working in a regime where string interactions exist.
Extending the simpler case of non-interacting strings to the regime where a black hole could exist requires supersymmetry. In certain cases, the entropy calculation done for zero string coupling remains valid when the strings interact. The challenge for a string theorist is to devise a situation in which a black hole can exist which does not "break" supersymmetry. In recent years, this has been done by building black holes out of D-branes. Calculating the entropies of these hypothetical holes gives results which agree with the expected Bekenstein entropy. Unfortunately, the cases studied so far all involve higher-dimensional spaces — D5-branes in nine-dimensional space, for example. They do not directly apply to the familiar case, the Schwarzschild black holes observed in our own universe.
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## History
Dirichlet boundary conditions and D-branes had a long `pre-history' before their full significance was recognized. Mixed Dirichlet/Neumann boundary conditions were first considered by Warren Siegel in 1976 as a means of lowering the critical dimension of open string theory from 26 or 10 to 4 (Siegel also cites unpublished work by Halpern, and a 1974 paper by Chodos and Thorn, but a reading of the latter paper shows that it is actually concerned with linear dilation backgrounds, not Dirichlet boundary conditions). This paper, though prescient, was little-noted in its time (a 1985 parody by Siegel, `The Super-g String,' contains an almost dead-on description of braneworlds). Dirichlet conditions for all coordinates including Euclidean time (defining what are now known as D-instantons) were introduced by Michael Green in 1977 as a means of introducing point-like structure into string theory, in an attempt to construct a string theory of the strong interaction. String compactifications studied by Harvey and Minahan, Ishibashi and Onogi, and Pradisi and Sagnotti in 1987-89 also employed Dirichlet boundary conditions.
The fact that T-duality interchanges the usual Neumann boundary conditions with Dirichlet boundary conditions was discovered independently by Horava and by Dai, Leigh, and Polchinski in 1989; this result implies that such boundary conditions must necessarily appear in regions of the moduli space of any open string theory. The Dai et al. paper also notes that the locus of the Dirichlet boundary conditions is dynamical, and coins the term Dirichlet-brane (D-brane) for the resulting object (this paper also coins orientifold for another object that arises under string T-duality). A 1989 paper by Leigh showed that D-brane dynamics are governed by the Dirac-Born-Infeld action. D-instantons were extensively studied by Green in the early 1990s, and were shown by Polchinski in 1994 to produce the e^{-1/g} nonperturbative string effects anticipated by Shenker. In 1995 Polchinski showed that D-branes are the sources of electric and magnetic Ramond–Ramond fields that are required by string duality, leading to rapid progress in the nonperturbative understanding of string theory.
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## See also
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## References
• Bachas, C. P. "Lectures on D-branes" (1998). arXiv:hep-th/9806199.
• Giveon, A. and Kutasov, D. "Brane dynamics and gauge theory," Rev. Mod. Phys. 71, 983 (1999). arXiv:hep-th/9802067.
• Hashimoto, Koji, D-Brane: Superstrings and New Perspective of Our World. Springer (2012). ISBN 978-3-642-23573-3
• Johnson, Clifford (2003). D-branes. Cambridge: Cambridge University Press. ISBN 0-521-80912-6.
• Polchinski, Joseph, TASI Lectures on D-branes, arXiv:hep-th/9611050. Lectures given at TASI '96.
• Polchinski, Joseph, Phys. Rev. Lett. 75, 4724 (1995). An article which established D-branes' significance in string theory.
• Zwiebach, Barton. A First Course in String Theory. Cambridge University Press (2004). ISBN 0-521-83143-1.
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http://math.stackexchange.com/questions/136848/fourier-transform-probability-distribution
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# Fourier transform probability distribution
Suppose I have 1,000 independent random values with a uniform distribution $[+1, -1]$. Now suppose I take the discrete Fourier transform of this data. What the heck is the probability distribution of the resulting Fourier coefficients?
(I'm guessing it's either uniformly distributed again, or else some manner of exotic distribution which converges to a normal distribution if the number of samples is high enough...)
Edit:
Each coefficient produced by the Fourier transform is the weighted sum of the input samples.
It appears that if $x$ and $y$ are both random variables, then the probability distribution of $x+y$ is equal to the convolution of the probability distribution for $x$ and the probability distribution for $y$. [Obviously, this applies iff $x$ and $y$ are independent.] I think I read that right, anyway!
If we were just summing $N$ I.I.D. variables, the result would be a normal distribution. But because it's a weighted sum, each weighted variable has a different [uniform] distribution. So... that means the answer is... uh...??
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May I ask discrete Fourier transform of what? So you have a vector $(a_1, ..., a_{1000}) \in [-1,1]^{1000}$ and you take the discrete Fourier transform of this vector? Or you bin the data, and take the discrete Fourier transform of the binned data? – Fabian Apr 25 '12 at 16:59
Just for the sake of curiosity, I tested in matlab what I interpreted to be the question: N=1e5;a=-1+2*rand(N,1);f=real(fftshift(fft(a)));histfit(f,50); and got what very much looks like a normal distribution (same for the the imaginary part) which, I suppose, is what the question is about. Could this somehow be related to the Box-Muller transform? – tibL Apr 25 '12 at 17:37
It does seem from Box-Muller that you'd be computing a weighted combination of normal components, and so the result ought to be normal. – EMS Apr 25 '12 at 22:29
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http://mathhelpforum.com/differential-geometry/143127-what-finite-subcover.html
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# Thread:
1. ## what is a finite subcover?
I'm trying to write up a proof and getting really stuck on this idea of "finite subcover."
The problem is to prove that given any closed and bounded interval [a,b] in R, any cover of [a,b] with open intervals contains a finite subcover.
I saw a proof by contradiction sketched out that went something like this:
Assume that there exists a cover of [a,b] that does not contain a finite subcover.
Divide the interval [a,b] in half. We know that one of those subintervals contains a cover that does not have a finite subcover. Divide in half again.... repeat process until we have a series of nested intervals all with the property that they are assumed to have an open cover that does not have a finite subcover.
The nested interval principle tells us that that the union of infinitly many nested subintervals will contain a single point. And here there is a contradiction that rests on an understanding of finite subcover. And that is where I'm stuck.
Can anyone please explain what a finite subcover is and why there is a contradition here? Is it that any open cover of a single point must have a finite subcover? Is that all it is?
2. A finite subcover is an open cover that is formed out of a finite number of sets. An open cover of E is a collection of open sets whose union contains E.
In fact, if E contain a finite number of element, then for any open cover of E, you can find a finite subcover. For example if E = {1,2,3}, then given any open cover (can be infinite of course), choose three sets from it in such a way that the first set contains 1, the second contains 2 and the third contain 3. You are sure that these three exist since we started from an open cover covering all E. Hence, the collection of the three sets found constitute a finite subcover. In you example, yes, any open cover of a single point must have a finite subcover since the given set contain a finite number of points (in fact only one point).
Hope this helps
3. Originally Posted by Beckeroo
I'm trying to write up a proof and getting really stuck on this idea of "finite subcover."
The problem is to prove that given any closed and bounded interval [a,b] in R, any cover of [a,b] with open intervals contains a finite subcover.
Suppose that $[a,b]$ is covered by a collection of open intervals.
Let $P=\{x\in (a,b]: [a,x]\text{ is covered by a finite subcollection.}\}$
1) Show that $P$ is not empty. Hint $a+\delta\in P$.
2) Let $q=\sup(P)$ then show that $q=b$. Hint what if $q<b?$
4. Originally Posted by mohammadfawaz
yes, any open cover of a single point must have a finite subcover since the given set contain a finite number of points (in fact only one point).
Hope this helps
I thought this must be the case but wasn't sure I was understanding correctly. Thanks!
5. Originally Posted by mohammadfawaz
A finite subcover is an open cover that is formed out of a finite number of sets.
Be careful here! Not just a "an open cover that is formed out of a finite number of sets". If that were true, since every set can be covered by a single open set, the entire space, every set would be compact! A finite open subcover is an open cover, consisting of a finite number of sets from a given open cover.
An open cover of E is a collection of open sets whose union contains E.
In fact, if E contain a finite number of element, then for any open cover of E, you can find a finite subcover. For example if E = {1,2,3}, then given any open cover (can be infinite of course), choose three sets from it in such a way that the first set contains 1, the second contains 2 and the third contain 3. You are sure that these three exist since we started from an open cover covering all E. Hence, the collection of the three sets found constitute a finite subcover. In you example, yes, any open cover of a single point must have a finite subcover since the given set contain a finite number of points (in fact only one point).
Hope this helps
6. Thank you HallsofIvy. I really appreciate your clarification.
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http://math.stackexchange.com/questions/266199/how-do-you-prove-this-set-problem?answertab=active
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# how do you prove this set problem?
I'm trying to teach myself set-theory. I have been unable to prove algebraically that:
$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B)$
I know it's elementary but I just want to see an example. (The book doesn't provide one.)
Correct me if I'm wrong...the first step is to use De Morgan's law to expand the left side. i.e.
$(A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap \overline{A} \cup \overline{B}$
$(A \cup B) \cap \overline{(A \cap B)} = A \cup (B \cap \overline{A}) \cup \overline{B}$, by associative property.
$(A \cup B) \cap \overline{(A \cap B)} = (A \cup \overline{A}) \cap (B \cup \overline{B} )$, by commutative property.
after that I don't know what to do.
I don't know whether I should rewrite these or if different notation would help. i.e. should I expand both sides to use the notations/definition of set-union? i.e. $A \cup B = \{x \mid x \in A \lor x \in B \}$
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## 5 Answers
$$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B)\tag{1}$$
Step (1) $(A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap \overline{A} \cup \overline{B}$
Yes, correct, but you should keep parentheses: $(A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap (\overline{A} \cup \overline{B})$
Step (2) Not correct: $(A \cup B) \cap \overline{(A \cap B)}$ = $A \cup (B \cap \overline{A}) \cup \overline{B}$, by associative property.
No: the associative property (like the commutative property) applies to a chain of unions, or a chain of intersections, but not a mixed chains of unions and intersections. This is why parentheses in the first step are needed. You need to use distribution:
$(A \cup B) \cap \overline{(A \cap B)}$ = $(A \cup B) \cap (\overline{A} \cup \overline{B}) = [(A\cup B) \cap \overline{A}] \cup [(A \cup B) \cap \overline{B}]$
Can you take it from here? You can use distributivity again...then use that fact that $A \cap \overline{A} = \varnothing$. Likewise for $B\cap \overline{B} = \varnothing$. Simplify.
Another approach is to unpack what proving your equality $(1)$ requires:
In general, we prove that for two sets $P, Q$, $$P = Q \iff P \subseteq Q \text{ AND}\;\;Q\subseteq P$$
For your equality $(1)$, that means you can prove the equality by proving: $$[(A \cup B) \cap \overline{(A \cap B)}] \subseteq [(A \cap \overline{B}) \cup (\overline{A} \cap B)]\tag{2}$$
and by proving $$[(A \cap \overline{B}) \cup (\overline{A} \cap B)] \subseteq [(A \cup B) \cap \overline{(A \cap B)}]\tag{3}$$
Unpack what this means in terms of "chasing elements", for which I'll give you a start:
$(2)$ If $x \in [(A \cup B)\cap \overline{(A \cap B)}]$ then $x \in (A\cup B)$ AND $x \notin (A \cap B)$, which means $(x \in A$ OR $x \in B)$ AND $(x \notin A$ OR $x \notin B)$...
$(3)$ If $x \in [(A\cap \overline{B})\cup (\overline{A} \cap B)]$, then ...$(x \in A \cap \overline{B})$ OR $(x \in \overline{A} \cap B)$ ....
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thanks that was perfect. – franklin Dec 29 '12 at 1:09
Your welcome, franklin. Glad to help! – amWhy Dec 29 '12 at 1:14
For future reference, the solution:
prove: $\ (A \cup B) \cap \overline{(A \cap B)} = (\overline{A} \cap B) \cup (\overline{B} \cap A)$
I will manipulate the left side only:
$\ (A \cup B) \cap (\overline{A} \cup \overline{B})$, use De Morgan's law
= $\ (( A \cup B) \cap \overline{A} ) \cup ((A \cup B) \cap \overline{B})$ , distribution
= $\ ( (\overline{A} \cap B) \cup (\overline{A} \cap A)) \cup ( (\overline{B} \cap B) \cup (\overline{B} \cap A))$ , distribution
remember that $\ \overline{A} \cap A = \emptyset$
= $\ (\overline{A} \cap B) \cup (\overline{B} \cap A)$
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Your first step is almost correct: you dropped the parentheses. Thus, you should have $$\begin{align}(A\cup B)\cap\overline{(A\cap B)}&=(A\cup B)\cap(\overline{A}\cup\overline{B})\\ &= \big[(A\cup B)\cap\overline{A}\big]\cup\big[(A\cup B)\cap\overline{B}\big]\\ &= (B\cap\overline{A})\cup(A\cap\overline{B}).\end{align}$$
As an example, we can take intervals on the real line, say $A=[0,2]$ and $B=[1,3]$. As an exercise, you can try to work it out to verify the identity as an example.
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Good first effort, but you don't have associativity and commutativity when dealing with the operations $\cup$ and $\cap$, so parentheses are super important.
It looks like you are missing what it means for two sets, call them $X$ and $Y$, to be equal. The definition is pretty simple: $X = Y$ means that any element $x \in X$ is also in $Y$, and every element $y \in Y$ is also contained in $X$.
Formally, $X = Y$ $\iff$ $X \subseteq Y$ and $Y \subseteq X$.
So you want to break this into two problems, showing that each is a subset of the other. So start out with "Let $x \in (A \cup B) \cap \overline{(A \cap B)}$..." and go from there. Let me know if you need more.
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$x \in A \wedge \not\in B \vee \not\in A \wedge \in B$
So x is in B and not A or in A but not B.
so x is in one of them but not both.
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http://cms.math.ca/10.4153/CMB-2011-063-8
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Canadian Mathematical Society
www.cms.math.ca
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# From Matrix to Operator Inequalities
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[PDF: 182KB]
http://dx.doi.org/10.4153/CMB-2011-063-8
Canad. Math. Bull. 55(2012), 339-350
Published:2011-04-06
Printed: Jun 2012
• Terry A. Loring,
Department of Mathematics and Statistics, University of New Mexico, Albuquerque, NM 87131, U.S.A.
Features coming soon:
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Format: HTML LaTeX MathJax PDF
## Abstract
We generalize Löwner's method for proving that matrix monotone functions are operator monotone. The relation $x\leq y$ on bounded operators is our model for a definition of $C^{*}$-relations being residually finite dimensional. Our main result is a meta-theorem about theorems involving relations on bounded operators. If we can show there are residually finite dimensional relations involved and verify a technical condition, then such a theorem will follow from its restriction to matrices. Applications are shown regarding norms of exponentials, the norms of commutators, and "positive" noncommutative $*$-polynomials.
Keywords: $C*$-algebras, matrices, bounded operators, relations, operator norm, order, commutator, exponential, residually finite dimensional
MSC Classifications: 46L05 - General theory of $C^*$-algebras 47B99 - None of the above, but in this section
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http://physics.stackexchange.com/questions/22398/direction-of-rotation-of-proton-in-magnetic-field-opposite-to-a-dipole/22402
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# Direction of rotation of proton in magnetic field--opposite to a dipole
Chatroom created by @pcr for discussing this: http://chat.stackexchange.com/rooms/2824/direction-of-rotation-of-proton-in-magnetic-field
Here's a small paradoxical question I was asked a long time ago (and have been asked twice since). I think I do know the answer, but I though it would be fun to ask it here.
Let's take a proton and fling it into a magnetic field coming out of the plane of the paper/screen ($\vec{B}=B_0\odot$) Now, looking from above the plane, the proton goes clockwise.
Alright. lets take our right hand and find the direction of the dipole moment. It's a proton going clockwise, so it's a clockwise current. This is a downwards ($\vec{M}=M_0\otimes$) dipole moment.
But, potential energy of a dipole is $U=-\vec{M}\cdot\vec{B}$. If they are antiparallel, then the dot product is negative, so we get $U=-M_0B_0(\odot\cdot\otimes)=M_0B_0$, and is positive. Compare that with the case where $\vec{M}\parallel\vec{B}$, we get a negative value of potential energy.
As we all know, a system tends to reduce its potential energy. Then why, in this case, does a proton deliberately choose the direction of rotation with the maximum potential energy?
## Reason for bounty
Multiple things. The bountybox does not provide the ability to overlap reasons, unfortunately ;)
• I have multiple conflicting answers, and while each one is individually convincing, when brought together the whole situation becomes a jumble
• I need more people to take a look at this, upvote answers they agree with, comment, and/or add more answers.
• The answers could be clearer
• It would help if the answers explained the paradox for various levels of understanding.
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I personally have two, simple explanations for this. What I want to see is how deep one can go into this.. And having been asked this question thrice, I guess it's appropriate for this site as it may be useful to future visitors. – Manishearth♦ Mar 15 '12 at 9:54
Atleast, I don't think so. If you can get another explanation from this, then please post it as an answer :). I have a feeling that this paradox can be explained in multiple ways, and some pretty deep. – Manishearth♦ Mar 15 '12 at 10:56
Don't ask me if I've thought of something or not. I already [think I] know the answer. I've posted this here to (a) Let others have fun; (b) see how deep one can go; (c) see how many explanations this has [most probably they'll all boil down to the same point]; (d) Put a [probably] common confusion/paradox on the site. If you feel you have an explanation, just post it as an answer. – Manishearth♦ Mar 15 '12 at 11:01
This question actually sounds familiar, I wonder if it's been asked here before? – David Zaslavsky♦ Mar 15 '12 at 18:34
@DavidZaslavsky: I asked it on chat once. No response. – Manishearth♦ Mar 16 '12 at 0:41
show 1 more comment
## 5 Answers
The potential energy in this case should be $U=+\vec{m}.\vec{B}$, hence the potential energy is minimized, as it should be. Here is the explanation:
Let’s look at the derivation of interaction energy between magnetic dipole and magnetic field carefully. The dipole energy $U=-\vec{m}.\vec{B}$ is derived using principle of virtual work with an assumption that the dipole moment is constant, and thus it’s self energy is permanent. However, if the dipole moment is allowed to change as in this case, the self energy of the dipole is no longer permanent. We can imagine it as the $\frac{1}{2}LI^2$ energy for the case of a current loop, if we change its dipole moment, its internal energy will also change. So for the case like this, the self energy can be extracted into mechanical energy. If we take into account the additional work to change the self energy in the principle of virtual work we will end up with $U=+\vec{m}.\vec{B}$. We can always calculate the work $\int \tau d\theta$ to change the orientation of the dipole. However in this case the dipole moment is not permanent, so its magnitude will be be different for different orientations. Thus the work calculation will be messy, but there is an easy way to do that. We can use some sort of "battery" to keep the dipole moment constant and calculate the work using $−\vec{m}.\vec{B}$. At the end of the process we put back all the energy given/stolen by the battery, which means I changed back the dipole moment to the value it should have been if the battery was not there. In other words I already get rid of all the influences of the battery. The work done by the battery turns out to be $(−2\vec{m}.\vec{B})$, then we get
$U=-\vec{m}.\vec{B}-(-2\vec{m}.\vec{B})=+\vec{m}.\vec{B}$
We can also get the same $U=+\vec{m}.\vec{B}$ if we calculate the total electromagnetic field energy, some details of the derivation is in my blog:
http://emitabsorb.wordpress.com/2011/08/21/m-b-or-m-b/
Magnetic field does no work on a proton, then how do we define potential energy?
Yes the total kinetic energy of the system is conserved, but we can separate it into parts. For example we can lump together the kinetic energy due to $v_x$ & $v_y$ and give it a name say $U_1$. The change in $U_1$ will affect the particle’s movement in $z$ direction, thus we can say that $U_1$ is the potential energy for $z$ direction. In this case we would like to know the tendency of the proton’s angular revolution velocity to align or counter-align with the magnetic field, so we lumped together part of the kinetic energy and magnetic field energy. As how it is derived, this energy can be written as $\tau=-dU/d\theta$. Thus if the lumped energy is not minimum, there will be torque perpendicular to $\vec{B}$.
So why don’t we also use $U=+\vec{m}.\vec{B}$ for the case of permanent dipole since it is the actual total energy with self energy already included in it?
Yes it is true that the right total energy is $U=+\vec{m}.\vec{B}$. But in this case the potential energy, the one that tends to minimize itself is $U=-\vec{m}.\vec{B}$. The part of energy that can minimize itself is the one that can be written as $F=-\nabla U$, that is to say the force will tend to any particle affected by the potential to the place where $U$ is lower. For example, consider a system of an earth and a moon orbiting it. Then suddenly the earth becomes twice as big as before with the same mass. We know that the self gravitational potential energy of the earth is changed, but it leaves no effect on the moon. So in this case the self gravitational potential energy of the earth is not a potential energy for the moon.
Now the only problem remaining in the permanent dipole case is that besides the $–\vec{m}.\vec{B}$ part which can change back and forth with mechanical energy, the remaining $+2\vec{m}.\vec{B}$(part of it is from the dipole’s self energy, and the remaining is from the self energy of the constant $\vec{B}$ field provider) part also changes mysteriously and which means the energy is not conserved. To save the principle conservation of energy we can always invent a new kind of energy so that the $+2\vec{m}.\vec{B}$ is not missing or being created freely but instead it is just changing its form between electromagnetic energy and this new energy. But I think it is not necessary, because what I was doing is not to protect the principle of conservation of energy, but instead to protect the field energy interpretation. Actually the field energy is also derived using the principle of virtual work in the first place, but in this case the increase in the total field energy is not equal to the decrease in mechanical energy. Thus for the case of permanent dipole I think the field energy interpretation is no longer valid. If we stick to the definition $F=-\nabla U$ , these difficulties would never occurs.
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Here is my comment from chat: "Chris' answer makes sense, but he doesn't address the real problem here. The fact that the proton radiates doesn't dictate the fact that it goes clockwise. We already know that the proton goes clockwise from basic EM law (or if you want, Lenz law). And now, we're trying to force the -m.B paradigm into our system -> which definitely won't work because we're not dealing with a permanent magnet." Radiation damping is necessary for 'full' description of the system, but that's for different question. – pcr Mar 18 '12 at 14:56
@Emitabsorb I think the battery argument works because in magnetostatics, we're dealing with a conservative system? Though it is no longer the case with the radiation included -> but again it's not the main issue, we can still explain the situation in the realm of magnetostatics – pcr Mar 18 '12 at 15:07
No we cannot, the problem is that you are trying to use a quasistatic approach without taking into account the full dynamics. And this view is ensured by my Berkeley colleagues. – Chris Gerig Mar 18 '12 at 22:52
– pcr Mar 18 '12 at 23:09
Yeah, this needs to be taken to the chat room. In the meantime I'll clean up the comments here. – David Zaslavsky♦ Mar 19 '12 at 7:55
show 6 more comments
I’m not an expert in electromagnetism, but if one of my students (general physics, BSc level) asked, I'd say the following: the rotating proton indeed generates an magnetic field, and it behaves like a magnetic dipole for these purposes (and at large enough distances). However, that only concerns the field created by this proton, and its interaction of this proton with other particles at long distance. The interaction energy between magnetic field and proton cannot be described as $-\vec M\cdot \vec B$, because that expression is the energy of a point dipole, which our system is not. It's like you would be calculating the torque of something that's not an rigid solid, it just doesn't apply.
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Here are my original resolutions to the paradox.
Looking at the other answer, this is most probably wrong, but I'll post it for completeness' sake. Also, it will emphasise what I suspected would happen: multiple resolutions to one paradox.
• Magnetic field does no work on a proton, thus there is no PE and the whole discussion is moot. Note that this may be wrong as the proton itself influences the local magnetic field.
• A proton doing UCM cannot be called a current loop. In a current loop, at any given instant, we have moving charges on every side. This gives us a couple. There is no such couple here.
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The argument of minimal potential energy is not applicable here, as it works only for conservative forces. Consider lagrangian $L=T(\dot q_i)-U(q_i)$ with $q_i$ being some generalised coordinates. One assumes normally, that any motion causes dissipative forces to diminish $\dot q_i$ with time, and hence the term $T(\dot q_i)$. Lagrange equations then read off $\dfrac{\partial U}{\partial q_i}=0$ and consideration of how the system approaches that point will show that it is a local mininum. By the way, the principle is particularly nice for being versatile with respect to the choice of $q_i$.
Now consider the proton. Its lagrangian function is $L=\dfrac{m v^2}{2}+\dfrac{e}{c}{\bf A}{\bf v}$. Potential energy here may be either:
1) Considered zero. Then any state is the minimal energy state.
2) Taken to be $-\dfrac{e}{c}{\bf A}{\bf v}$. Then the system is not conservative and the argument doesn't apply.
Hence, the energy argument $U\rightarrow U_\min$ doesn't say anything about the state where the system shall be.
If you switch to magnetic momentum formulation, you use lagrangian $L=\dfrac{m v^2}{2}+{\bf m}{\bf H}$. Implicit assumption about ${\bf m}$ being fixed makes one expect that after enough disipation the system would reach ${\bf m}{\bf H}=-|{\bf m}{\bf H}|$. Instead, as seen from lagrangian, the system would just trivially reach the state ${\bf v}=0$ and hence ${\bf m}=0$.
To conclude, the paradox stems from applying $U\rightarrow U_\min$ and assuming ${\bf m}$ constant for dipole discription.
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I think this is slightly mudddling the point, and is an attempted restatement of my answer. Your Lagrangian does not take into consideration radiation damping (in this case being cyclotron radiation), which is what my post is all about (and this is equivalent to "nonconservative force"). – Chris Gerig Mar 20 '12 at 20:40
Indeed, you are making use of your Lagrangian and showing the the principle of minimum (total) energy does not hold here... But that is clear because the system is not closed, there is radiation (as I have stated). The real principle going on here is the minimum potential energy principle which holds for ALL systems, and that is what my post addresses. – Chris Gerig Mar 20 '12 at 20:50
Thank you, Chris, but: 1) Radiation damping is a dissipative force and hence cannot be used in lagrangian, 2) Were there no radiation damping, the paradox would still hold, possibly in a modified form, involving statistics – Alexey Bobrick Mar 20 '12 at 20:54
Chris, your post is great, I don't claim that you are wrong, or not original, but I just give a simple rigorous explanation of what is going on. Besides, the minimal potential energy principle doesn't hold for "ALL" systems, you are wrong here. – Alexey Bobrick Mar 20 '12 at 20:57
And actually, I take it back, your argument is simply wrong because $-A\cdot v$ IS a generalized conservative potential energy. The fact is, you have not considered radiation. – Chris Gerig Mar 20 '12 at 20:57
show 5 more comments
1) The answer of F'x is incorrect, there is definitely a magnetic dipole moment (and there is no self-energy here of its moment with its own magnetic field), and it indeed points downward. However, the proton accelerates in its cyclotron movement, inducing radiation and hence depleting the energy, reconciling this paradox.
2) Alexey's answer is flawed because his Lagrangian does not take into consideration the full dynamics. He also argues that his Lagrangian system is not conserved (but it is in his post).
3) They key response to Emitabsorb's answer:
$U=-m\cdot B$ is the interaction potential from a moment in a field; it only includes the work done in establishing an a priori permanent magnetic moment in the field. It does not include the work done in creating the magnetic moment and keeping it permanent, however. If we include this extra work, the total energy of the system becomes $U=+\bar{m}\cdot \bar{B}$. But you still want to minimize the interaction-potential only (this is explained in Jackson's Classical EM textbook, pg190 + pg214). HOWEVER, we assumed here that you first had a moment $m$ independent of any magnetic fields, and then brought it into $B$. This simply doesn't apply to our scenario... if you write down the Lagrangian for a proton in a magnetic field, there is no interaction potential $\pm m\cdot B$ that appears (because there is no a priori fixed moment); it is simply $\mathcal{L}=\frac{1}{2}m_0v^2-\frac{e}{c}\bar{v}\cdot \bar{A}$. This paragraph, coupled with my acceleration-remark, resolves the paradox!!!
My acceleration-remark:
In effect, you're trying to consider a quasistatic solution to this problem, without taking into consideration the full dynamics. The proton circles around and produces cyclotron radiation, damping its motion (i.e. radiation damping). You can form a Lagrangian to handle this, and minimum potential energy is achieved.
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## protected by Qmechanic♦Jan 23 at 17:15
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://mathoverflow.net/questions/103279/isoperimetry-and-poincare-inequality
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## Isoperimetry and Poincare Inequality
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What are the known relations between isoperimetric and Poincare inequalities on manifolds?
For example, for manifolds with a lower bound on Ricci curvature, the Cheeger-Buser inequality relates the isoperimetric constant to the first eigenvalue of the Laplacian. Does this imply a Poincare inequality for such manifolds?
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## 1 Answer
Everything here is for closed Riemannian manifolds. If you have a lower bound on Cheeger's isoperimetric constant $h(M)$, then Cheeger's inequality $\lambda_1(M)\geq \frac{h(M)^2}{4}$ gives you a lower bound on the first eigenvalue of the Laplacian. Thanks to the variational characterization of $\lambda_1(M)$, this is exactly equivalent to an upper bound on the constant $C_P$ in the $L^2$ Poincaré inequality $$\int_M f^2 dV \leq C_P \int_M |\nabla f|^2 dV,$$ for all smooth functions $f$ with $\int_M f dV=0$.
If you have a lower bound for the Ricci curvature, and also an upper bound for $C_P$ (equivalently, a lower bound for $\lambda_1(M)$), then Buser's inequality gives you a lower bound for $h(M)$.
See for example Wikipedia's page.
On the other hand a result of Yau shows that $h(M)$ is equal to the reciprocal of the constant $C_P'$ in the $L^1$ Poincaré inequality $$\int_M |f| dV \leq C_P' \int_M |\nabla f| dV,$$ for all smooth functions $f$ with $\int_M f dV=0$.
If you are interested in obtaining bounds on $h(M)$ or $\lambda_1(M)$ in terms of geometric quantities, there are the following basic results: $h(M)$ can be bounded below in terms of a lower bound for the volume of $M$, an upper bound for its diameter, and a lower bound for its Ricci curvature (Croke). On the other hand $\lambda_1(M)$ can be bounded below in terms of an upper bound for its diameter and a lower bound for its Ricci curvature (Li-Yau). Of course these bounds also depend on the dimension of the manifold.
Finally, if you are intersted in the relation with Sobolev inequalities, you should look at this paper of Peter Li.
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A really good book on this is Chavel's "Eigenvalues in Riemannian Geometry" – Deane Yang Jul 27 at 15:39
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http://mathoverflow.net/questions/10870?sort=votes
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## Which topological spaces admit a nonstandard metric?
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My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R* in place of the usual R-valued metric.
That is, let us define that a topological space X is a nonstandard metric space, if there is a distance function, not into the reals R, but into some nonstandard R* in the sense of nonstandard analysis. That is, there should be a distance function d from X2 into R*, such that d(x,y)=0 iff x=y, d(x,y)=d(y,x) and d(x,z) <= d(x,y)+d(y,z). Such a nonstandard metric would give rise to the nonstandard open balls, which would generate a metric-like topology on X.
There are numerous examples of such spaces, beginning with R* itself. Indeed, every metric space Y has a nonstandard analogue Y*, which is a nonstandard metric space. In addition, there are nonstandard metric spaces that do not arise as Y* for any metric space Y. Most of these examples will not be metrizable, since we may assume that R* has uncountable cofinality (every countable set is bounded), and this will usually prevent the existence of a countable local basis. That is, the nested sequence of balls around a given point will include the balls of infinitesimal radius, and the intersection of any countably many will still be bounded away from 0. For example, R* itself will not be metrizable. The space R* is not connected, since it is the union of the infinitesimal neighborhoods of each point. In fact, one can show it is totally disconnected.
Nevertheless, it appears to me that these nonstandard metric spaces are as useful in many ways as their standard counterparts. After all, one can still reason about open balls, with the triangle inequality and whatnot. It's just that the distances might be nonstandard. What's more, the nonstandard reals offer some opportunities for new topological constructions: in a nonstandard metric space, one has the standard-part operation, which would be a kind of open-closure of a set---For any set Y, let Y+ be all points infinitesimally close to a point in Y. This is something like the closure of Y, except that Y+ is open! But we have Y subset Y+, and Y++=Y+, and + respects unions, etc.
In classical topology, we have various metrization theorems, such as Urysohn's theorem that any second-countable regular Hausdorff space is metrizable.
Question. Which toplogical spaces admit a nonstandard metric? Do any of the classical metrization theorems admit an analogue for nonstandard metrizability? How can we tell if a given topological space admits a nonstandard metric?
I would also be keen to learn of other interesting aspects of nonstandard metrizability, or to hear of interesting applications.
I have many related questions, such as when is a nonstandard metric space actually metrizable? Is there any version of R* itself which is metrizable? (e.g. dropping the uncountable cofinality hypothesis)
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I think what you're talking about can be expressed more naturally in terms of non-metrizable uniform spaces. – Harry Gindi Jan 6 2010 at 2:32
Well, every nonstandard metric space is uniform for the same reason that every metric space is uniform. But are you proposing it as an equivalence? – Joel David Hamkins Jan 6 2010 at 3:17
No, I was just saying that *R-metrizability can be reduced to a much nicer condition in terms of standard analysis by using uniform spaces. – Harry Gindi Jan 6 2010 at 4:16
It seems that Dorais's answer bears out your expectation! So thanks very much. Although I suppose "nice" is relative; after all, the R* metric captures in one distance function the same information as uncountably many pseudo-metrics... – Joel David Hamkins Jan 6 2010 at 5:22
True, but we understand the standard reals much better than the nonstandard reals. – Harry Gindi Jan 6 2010 at 10:17
show 1 more comment
## 4 Answers
The uniformity defined by a *R-valued metric is of a special kind.
Let $(n_i)_{i<\kappa}$ be a cofinal sequence of positive elements in *R. We may assume that $i < j$ implies that $n_i/n_j$ is infinitesimal.
Given a *R-valued metric space $(X,d)$ we can define a family $(d_i)_{i<\kappa}$ of pseudometrics by $d_i(x,y) = st(b(n_id(x,y)))$, where $st$ is the standard part function and $b(z) = z/(1+z)$ to make the metrics bounded by $1$. The topology on $X$ is defined by the family of pseudometrics $(d_i)_{i<\kappa}$. Note that these pseudometrics have a the following special property:
(+) If $i < j < \kappa$ and $d_j(x,y) < 1$ then $d_i(x,y) = 0$.
Every uniform space can be defined by a family of pseudometrics, but it is rather unusual for the family to have property (+) when $\kappa > \omega$.
On the other hand, given a family of pseudometrics $(d_i)_{i<\kappa}$ bounded by $1$ with property (+), then we can recover a *R-valued metric by defining $d(x,y) = d_i(x,y)/n_i$, where $i$ is minimal such that $d_i(x,y) > 0$.
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Wow, that is great! Thanks very much. – Joel David Hamkins Jan 6 2010 at 4:26
I'm guessing that (+) is equivalent to the uniformity being generated by a fundamental system of size $\kappa$ which is moreover $<\kappa$-directed, but that's just a guess. – François G. Dorais♦ Jan 6 2010 at 4:46
I don't see that you ever use your assumption that kappa is uncountable. – Joel David Hamkins Jan 6 2010 at 4:52
Yes, I just fixed that, thanks. (Those were remnants of an earlier more complicated answer.) – François G. Dorais♦ Jan 6 2010 at 5:13
Your sequence of pseudo metrics seems similar to a representation of d in an ultrapower presentation of R*. That is, if R* arises as an ultrapower of R, then can one use the ultrapower representations directly to build the pseudometrics? – Joel David Hamkins Jan 6 2010 at 14:34
show 4 more comments
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
These are $\omega_\mu$-metrizable spaces, where $\omega_\mu$ is the cofinality of ${}^*\mathbb{R}$. Take a decreasing sequence $\langle x_\alpha:\alpha<\omega_\mu\rangle$ of positive elements that converges to $0$; because $\omega_\mu$ is regular uncountable one can ensure that $x_{\alpha+1}/x_\alpha$ is always infinitesimal. Any `metric' $d:X\times X\to\mathbb{R}$ can be converted to an ultrametric $\rho$ with values in ${x_\alpha:\alpha<\omega_\mu}$; it satisfies the strong triangle inequality: $\rho(x,z)\le\max{\rho(x,y),\rho(y,z)}$. These spaces are also called linearly uniformizable: the sets $A_\alpha=\lbrace (x,y):\rho(x,y) < x_\alpha \rbrace$ form a (linearly ordered) base for a uniformity that generates the same topology as $\rho$. Also, because of the uncountable cofinality, these spaces are $P$-spaces: $G_\delta$-sets are open. The only metrizable such spaces must be discrete. Here is one of the earliest papers that I am aware of to treat this kind thing systematically.
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I don' t know how relevant this is to your question, but some topologists have investigated quite extensively topological spaces with metrics in a lattice-ordered group, ie a group with a lattice order, compatible with the group operation. Your R*-metric spaces seem to be a chapter of this general theory (and similarly the theory of spaces with metrics in ON, which maybe are also of interest to you), so perhaps a few questions you have touched upon in your post can be answered from a broader perspective.
Here is an interesting paper by several authors, one of whom is Ralph Kopperman:
http://www.wku.edu/~tom.richmond/Papers/l-Groups.pdf
PS Kopperman has two desiderable features from your geolocation's standpoint, the first is working in the NYC area (CCNY if I am not mistaken), and the second one is being a friend and collaborator of Prabhud Misra (and thus, via the small world law, very reachable). If I remember well, Ralph is an expert in these type of things, and so you may ask him for further infos, if needs be.
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1
Oh yes, of course I've met Ralph and have occassionally attended the traveling seminar he runs with Misra, whom I know very well. Thanks for this answer. – Joel David Hamkins Jul 7 at 22:25
I suspected so. Anyway, thank you for the very charming question. Incidentally, here is another interesting (I think) viewpoint: some folks in cat theory have studied topoi in the context of synthetic differential geometry, where (inside the topos) there is one fellow whom the topos believes to be R, but which in fact has real infinitesimals. Now, If one considers a "metric space" object inside this topos, it would be one of your gadgets. I am sure you can play a similar trick inside your beloved ZFC models, where the "R" is (from outside) R*. Anyone has tried something along these lines? – Mirco Mannucci Jul 7 at 22:42
I once considered the following situation: $[0;1]^X$ is metrizable (with values in $\mathbb{R}$), if $X$ is countable. So I wondered whether $[0;1]^\mathbb{R}$ is nonstandard-metrizable for some totally ordered field $F$.
However it turns out that it does not admit a totally ordered local basis, so it can't be nonstandard-metrizable. I could give more details, if somebody is interested in them.
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http://math.stackexchange.com/questions/218137/the-necessary-and-sufficient-condition-for-diffeomorphism/218213
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# The necessary and sufficient condition for diffeomorphism
I came to the following proposition:
Let $M$ and $N$ be two smooth manifolds with respective maximal atlases $\mathcal{A}_M,\mathcal{A}_N$. Then a bijection $f:M \rightarrow N$ is a diffeomorphism if and only if the following codition holds:
$$(U,y)\in\mathcal{A}_N \Leftrightarrow (f^{-1}(U), y\circ f)\in\mathcal{A}_M$$
I can prove the necessary part, but don't know how to prove the sufficiency.
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## 1 Answer
We assume that $\mathcal{A}_{M}$ and $\mathcal{A}_{N}$ are maximal $C^{\infty}$-atlases on $M$ and $N$ respectively. The proof of sufficiency then proceeds as follows.
The chart maps coming from $\mathcal{A}_{M}$ and $\mathcal{A}_{N}$ are automatically diffeomorphisms (to prove this, use the fact that transition functions are smooth by definition). Hence, $y: U \rightarrow y[U]$ and $y \circ f: {f^{-1}}[U] \rightarrow y[U]$ are both diffeomorphisms. It thus follows that both $$f|_{{f^{-1}}[U]} = y^{-1} \circ (y \circ f): {f^{-1}}[U] \rightarrow U$$ and $$f^{-1}|_{U} = (y \circ f)^{-1} \circ y: U \rightarrow {f^{-1}}[U]$$ are diffeomorphisms, hence smooth functions.
Observe that $\{ {f^{-1}}[U] \,|\, (U,y) \in \mathcal{A}_{N} \}$ is an open cover of $M$ and that $f$ is smooth on each piece of the cover. Hence, by the smooth version of the Pasting Lemma, $f$ is smooth globally.
Likewise, $\{ U \,|\, (U,y) \in \mathcal{A}_{N} \}$ is an open cover of $N$ and $f^{-1}|_{U}$ is smooth on each piece of the cover. Hence, $f^{-1}$ is smooth globally.
We therefore conclude that $f: M \rightarrow N$ is a diffeomorphism.
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Why does chart map need to be diffeomorphism? By definition, they are just bijections between $\mathcal{M}$ and $\mathcal{R}^n$. You hints 'transition functions', but I don't understand. What if there is only one global chart? Then there will be no transition function. – hxhxhx88 Oct 22 '12 at 2:25
Still a little confused. Diffeomorphism is a concept in the smooth manifold category, I think it has nothing to do with a chart map. Actually, your proof give me an insight. By definition, when $f:M\rightarrow N$ satisfies both $\phi\circ f\circ\varphi^{-1}$ and $\varphi\circ f^{-1}\circ\phi^{-1}$ are smooth where $\varphi$ and $\phi$ are charts of $M$ and $N$, it is a diffeomorphism. So we only need to check $y\circ f\circ(y\circ f)^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n$ and $(y\circ f)\circ f^{-1}\circ y^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n$ to be smooth. – hxhxhx88 Oct 23 '12 at 5:05
Obviously they are, because they are identity.. So I think we do not need the chart map to be smooth..I think maybe we are not agree on the definition of diffeomorphism:) – hxhxhx88 Oct 23 '12 at 5:05
I think we don't need to check all pairs $(\varphi, \phi)$, because by definition for every point $P$ in $M$, there should exists a neighborhood on which $f$ is smooth. Because $f$ in a bijection as assumed, I only need to pick $f^{-1}(U)$ and $U$ with corresponding chart maps. Still don't understand your words: yes, when defining a $C^{\infty}$-atlas, we need $\phi \circ \psi^{-1}$ to be smooth, but it does not imply and require $\phi$ and $\psi$ themselves to be smooth. I'm new to this subject, hope I'm not making your mad :) – hxhxhx88 Oct 23 '12 at 5:34
Yeah, but it doesn't mean $\varphi$ itself should be smooth, we just require the composition $\varphi_1\circ\varphi_2^{-1}$ to be smooth, right?...Actually, what does the smoothness of a chart map mean?.. – hxhxhx88 Oct 23 '12 at 5:43
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http://mathforum.org/mathimages/index.php?title=Solving_Triangles&diff=23733&oldid=23732
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# Solving Triangles
### From Math Images
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| - | The main image and the details about it come from http://www.imdb.com/title/tt0105378/. | | |
| | Some of the ideas for problems/pictures on this page are based from ideas or concepts in the ''Interactive Mathematics Program'' Textbooks by Fendel, Resek, Alper and Fraser. | | Some of the ideas for problems/pictures on this page are based from ideas or concepts in the ''Interactive Mathematics Program'' Textbooks by Fendel, Resek, Alper and Fraser. |
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## Revision as of 11:57, 7 July 2011
The Shadow Problem
Field: Geometry
Created By: Orion Pictures
Website: [ http://en.wikipedia.org/wiki/File:Shadows_and_fog.jpg]
The Shadow Problem
In the 1991 film Shadows and Fog, the eerie shadow of a larger-than-life figure appears against the wall as the shady figure lurks around the corner. How tall is the ominous character really? Filmmakers use the geometry of shadows and triangles to make this special effect.
The shadow problem is a standard type of problem for teaching trigonometry and the geometry of triangles. In the standard shadow problem, several elements of a triangle will be given. The process by which the rest of the elements are found is referred to as solving a triangle.
# Basic Description
A triangle has six total elements: three sides and three angles. Sides are valued by length, and angles are valued by degree or radian measure. According to postulates for congruent triangles, given three elements, other elements can always be determined as long as at least one side length is given. Math problems that involve solving triangles, like shadow problems, typically provide certain information about just a few of the elements of a triangle, so that a variety of methods can be used to solve the triangle.
Shadow problems normally have a particular format. Some light source, often the sun, shines down at a given angle of elevation. The angle of elevation is the smallest—always acute—numerical angle measure that can be measured by swinging from the horizon from which the light source shines. Assuming that the horizon is parallel to the surface on which the light is shining, the angle of elevation is always equal to the angle of depression. The angle of depression is the angle at which the light shines down, compared to the angle of elevation which is the angle at which someone or something must look up to see the light source. Knowing the angle of elevation or depression can be helpful because trigonometry can be used to relate angle and side lengths.
In the typical shadow problem, the light shines down on an object or person of a given height. It casts a shadow on the ground below, so that the farthest tip of the shadow makes a direct line with the tallest point of the person or object and the light source. The line that directly connects the tip of the shadow and the tallest point of the object that casts the shadow can be viewed as the hypotenuse of a triangle. The length from the tip of the shadow to the point on the surface where the object stands can be viewed as the first leg, or base, of the triangle, and the height of the object can be viewed as the second leg of the triangle. In the most simple shadow problems, the triangle is a right triangle because the object stands perpendicular to the ground.
In the picture below, the sun casts a shadow on the man. The length of the shadow is the base of the triangle, the height of the man is the height of the triangle, and the length from the tip of the shadow to top of the man's head is the hypotenuse. The resulting triangle is a right triangle.
In another version of the shadow problem, the light source shines from the same surface on which the object or person stands. In this case the shadow is projected onto some wall or vertical surface, which is typically perpendicular to the first surface. In this situation, the line that connects the light source, the top of the object and the tip of the shadow on the wall is the hypotenuse. The height of the triangle is the length of the shadow on the wall, and the distance from the light source to the base of the wall can be viewed as the other leg other leg of the triangle. The picture below diagrams this type of shadow problem, and this page's main picture is an example of one of these types of shadows.
More difficult shadow problems will often involve a surface that is not level, like a hill. The person standing on the hill does not stand perpendicular to the surface of the ground, so the resulting triangle is not a right triangle. Other shadow problems may fix the light source, like a street lamp, at a given height. This scenario creates a set of two similar triangles.
Ultimately, a shadow problem asks you to solve a triangle given only a few elements of the possible six total. In the case of some shadow problems, like the one that involves two similar triangles, information about one triangle may be given and the question may ask to find elements of another.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Trigonometry, Geometry
[Click to view A More Mathematical Explanation]
## Why Shadows?
Shadows are useful in the set-up of a triangle pro [...]
[Click to hide A More Mathematical Explanation]
## Why Shadows?
Shadows are useful in the set-up of a triangle problems because of the way light works. A shadow is cast when light cannot shine through a solid surface. Light shines in a linear fashion, that is to say it does not bend. Light waves travel forward in the same direction in which the light was shined.
In addition to the linear fashion in which light shines, light has certain angular properties. When light shines on an object that reflects light, it reflects back at the same angle at which it shined. Say a light shines onto a mirror. The angle between the beam of light and the wall that the mirror is the angle of approach. The angle from the wall at which the light reflects off of the mirror is the angle of departure. The angle of approach is equal to the angle of departure.
Light behaves the same way a cue ball does when it is bounced off of the wall of a pool table at a certain angle. Just like the way that the cue ball bounces off the wall, light reflects off of the mirror at exactly the same angle at which it shines. The beam of light has the same properties as the cue ball in this case: the angle of departure is the same as the angle of approach. This property will help with certain types of triangle problems, particularly those that involve mirrors.
## More Than Just Shadows
Shadow problems are just one type of problem that involves solving triangles. There are numerous other formats and set ups for unsolved triangle problems. Most of these problems are formatted as word problems; they set up a triangle problem in terms of some real life scenario.
There are, however, many problems that simply provide numbers that represent angles and side lengths. In this type of problem, angles are denoted with capital letters, ${A, B, C,...}$, and the sides are denoted by lower-case letters,${a,b,c,...}$, where $a$ is the side opposite the angle $A$.
Ladder Problems One other common problem in solving triangles is the ladder problem. A ladder of a given length is leaned up against a wall that stands perpendicular to the ground. The ladder can be adjusted so that the top of the ladder sits higher or lower on the wall and the angle that the ladder makes with the ground increases or decreases accordingly. Because the ground and the wall are perpendicular to one another, the triangles that need to be solved in ladder problems always have right angles. Since the right angle is always fixed, many ladder problems require the angle between the ground and the ladder, or the angle of elevation, to to be somehow associated with a fixed length of a ladder and the height of the ladder on the wall. In other words, ladder problems normally ask for the height of the ladder on the wall or the ground distance between the ladder and the wall, and typically require some trigonometric calculation.
Mirror Problems Mirror problems are a specific type of triangle problem which involves two people or objects that stand looking into the same mirror. Because of the way a mirror works, light reflects back at the same angle at which it shines in, as explained below in A More Mathematical Explanation. In a mirror problem, the angle at which one person looks into the mirror, or the angle of vision is the same exact angle at which the second person must look into the mirror to make eye contact. Typically, the angle at which one person looks into the mirror is given along with some other piece of information. Once that angle is known, then one angle of the triangle is automatically known since the light reflects back off of the mirror at the same angle, making the angle of the triangle next to the mirror the supplement to twice the angle of vision.
Sight Problems Like shadow problems, sight problems include many different scenarios and several forms of triangles. Most sight problems are set up as word problems. They involve a person standing below or above some other person or object. In most of these problems, a person measures an angle with a tool called an astrolabe or a protractor. In the most standard type of problem, a person uses the astrolabe to measure the angle at which he looks up or down at something. In the example at the right, the bear stands in a tower of a given height and uses the astrolabe to measure the angle at which he looks down at the forest fire. The problem asks to find how far away the forest fire is from the base of the tower given the previous information.
## Ways to Solve Triangles
In all cases, a triangle problem will only give a few elements of a triangle and will ask to find one or more of the lengths or angle measures that is not given. There are numerous formulas, methods, and operations that can help to solve a triangle depending on the information given in the problem.
The first step in any triangle problem is drawing a diagram. A picture can help to show which elements of the triangle are given and which elements are adjacent or opposite one another. By knowing where the elements are in relation to one another, we can use the trigonometric functions to relate angle and side lengths.
There are many techniques which can be implemented in solving triangles:
• Trigonometry: The basic trigonometric functions relate side lengths to angles. By substituting the appropriate values into the formulas for sine, cosine, or tangent, trigonometry can help to solve for a particular side length or angle measure of a right triangle. This is useful when given a side length and an angle measure.
• Inverse Trigonometry: Provided two side lengths, the inverse trig functions use the ratio of the two lengths and output an angle measure in right triangle trigonometry. Inverse trig is particularly useful in finding an angle measure when two side lengths are given in a right triangle.
• Special Right Triangles: Special right triangles are right triangles whose side lengths produce a particular ratio in trigonometry. A 30°− 60°− 90° triangle has a hypotenuse that is twice as long as one of its legs. A 45°− 45°− 90° is called an isosceles right triangle since both of its legs are the same length. These special cases can help to quicken the process of solving triangles.
• Pythagorean Theorem: The Pythagorean Theorem relates the squares of all three side lengths to one another in right triangles. This is useful when a triangle problem provides two side lengths and a third is needed.
$a^{2}+b^{2} = c^{2}$
• Pythagorean Triples: A Pythagorean triple is a set of three positive integers that satisfy the Pythagorean Theorem. The set {3,4,5} is one of the most commonly seen triples. Given a right triangle with legs of length 3 and 4, for example, the hypotenuse is known to be 5 by Pythagorean triples.
• Law of Cosines: The law of cosines is a generalization of the Pythagorean Theorem which can be used for solving non-right triangles. The law of cosines relates the squares of the side lengths to the cosine of one of the angle measures. This is particularly useful given a SAS configuration, or when three side lengths are known and no angles for non-right triangles.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
• Law of Sines: The law of sines is a formula that relates the sine of a given angle to the length of its opposite side. The law of sines is useful in any configuration when an angle measure and the length of its opposite side are given. It is also useful given an ASA configuration, and often the ASS configuration . The ASS configuration is known as the ambiguous case since it does not always provide one definite solution to the triangle.
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
When solving a triangle, one side length must always be given in the problem. Given an AAA configuration, there is no way to prove congruency. According to postulates for congruent triangles, the AAA configuration proves similarity in triangles, but there is no way to find the side lengths of a triangle. Knowing just angle measures is not helpful in solving triangles.
## Example Triangle Problems
Example 1: Using Trigonometry
A damsel in distress awaits her rescue from the tallest tower of the castle. A brave knight is on the way. He can see the castle in the distance and starts to plan his rescue, but he needs to know the height of the tower so he can plan properly. The knight sits on his horse 500 feet away from the castle. He uses his handy protractor to find the measure of the angle at which he looks up to see the princess in the tower, which is 15°. Sitting on the horse, the knight's eye level is 8 feet above the ground. What is the height of the tower?
We can use tangent to solve this problem. For a more in depth look at tangent, see Basic Trigonometric Functions.
Use the definition of tangent.
$\tan =\frac{\text{opposite}}{\text{adjacent}}$
Plug in the angle and the known side length.
$\tan 15^\circ =\frac{x \text{ft}}{500 \text{ft}}$
Clearing the fraction gives us
$\tan 15^\circ (500) =x$
Simplify for
$(.26795)(500) =x$
Round to get
$134 \text{ft} \approx x$
But this is only the height of the triangle and not the height of the tower. We need to add 8 ft to account for the height between the ground and the knight's eye-level which served as the base of the triangle.
$134 \text{ft} + 8 \text{ft} = h$
simplifying gives us
$142 \text{ft} = h$
The tower is approximately 142 feet tall.
Example 2: Using Law of Sines
A man stands 100 feet above the sea on top of a cliff. The captain of a white-sailed ship looks up at a 45° angle to see the man, and the captain of a black-sailed ship looks up at a 30° angle to see him. How far apart are the two ships?
To solve this problem, we can use the law of sines to solve for the bases of the two triangles since we have an AAS configuration with a known right angle. To find the distance between the two ships, we can take the difference in length between the bases of the two triangles.
First, we need to find the third angle for both of the triangles. Then we can use the law of sines.
For the white-sailed ship,
$180^\circ - 90^\circ - 45^\circ = 45^\circ$
Let the distance between this ship and the cliff be denoted by $a$.
By the law of sines,
$\frac{100}{\sin 45^\circ} = \frac{a}{\sin 45^\circ}$
Multiplying both sides by $\sin 45^\circ$ gives us
$(\sin 45^\circ)\frac{100}{\sin 45^\circ} = a$
Simplify for
$a = 100 \text{ft}$
For the black-sailed ship,
$180^\circ - 90^\circ - 30^\circ =60^\circ$
Let the distance between this ship and the cliff be denoted by $b$.
By the law of sines,
$\frac{100}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$
Clear the fractions to get,
$100(\sin 60^\circ) = b(\sin 30^\circ)$
Compute the sines of the angle to give us
$100\frac{\sqrt{3}}{2} = b\frac{1}{2}$
Simplify for
$100(\sqrt{3}) = b$
Multiply and round for
$b =173 \text{ft}$
The distance between the two ships, $x$, is the positive difference between the lengths of the bases of the triangle.
$b-a=x$
$173-100 = 73 \text{ft}$
The ships are about 73 feet apart from one another.
Example 3: Using Multiple Methods
At the park one afternoon, a tree casts a shadow on the lawn. A man stands at the edge of the shadow and wants to know the angle at which the sun shines down on the tree. If the tree is 51 feet tall and if he stands 68 feet away from the tree, what is the angle of elevation?
There are several ways to solve this problem. The following solution uses a combination of the methods described above.
First, we can use Pythagorean Theorem to find the length of the hypotenuse of the triangle, from the tip of the shadow to the top of the tree.
$a^{2}+b^{2} = c^{2}$
Substitute the length of the legs of the triangle for $a, b$
$51^{2}+68^{2} = c^{2}$
Simplifying gives us
$2601+4624 = c^{2}$
$7225 = c^{2}$
Take the square root of both sides for
$\sqrt{7225} = c$
$85 = c$
Next, we can use the law of cosines to find the measure of the angle of elevation.
$a^{2}=b^{2}+c^{2} - 2bc \cos A$
Plugging in the appropriate values gives us
$51^{2}=68^{2}+85^{2} - 2(85)(68) \cos A$
Computing the squares gives us
$2601= 4624+7225 - 11560 \cos A$
Simplify for
$2601= 11849 - 11560 \cos A$
Subtract $11849$ from both sides for
$-9248= -11560 \cos A$
Simplify to get
$.8 = \cos A$
Use inverse trigonometry to find the angle of elevation.
$A = 37^\circ$
# Why It's Interesting
Shadow Problems are one of the most common types of problems used in teaching trigonometry. A shadow problem sets up a scenario that is simple, visual, and easy to remember. Shadow problems are commonly used and highly applicable.
Shadows, while an effective paradigm in a word problem, can even be useful in real life applications. In this section, we can use real life examples of using shadows and triangles to calculate heights and distances.
## Example: Sizing Up Swarthmore
The Clothier Bell Tower is the tallest building on Swarthmore College's campus, yet few people know exactly how tall the tower stands. We can use shadows to determine the height of the tower. Here's how:
Step 1) Mark the shadow of the of the tower. Make sure to mark the time of day. The sun is at different heights throughout the day. The shadows are longest earlier in the morning and later in the afternoon. At around midday, the shadows aren't very long, so it might be harder to find a good shadow. When we marked the shadow of the bell tower, it was around 3:40 pm in mid-June.
Step 2) After marking the shadow, we can measure the distance from our mark to the bottom of the tower. This length will serve as the base of our triangle. In this case, the length of the shadow was 111 feet.
Step 3) Measure the angle of the sun at that time of day.
Use a yardstick to make a smaller, more manageable triangle. Because the sun shines down at the same angle as it does on the bell tower, the small triangle and the bell tower's triangle are similar and therefore have the same trigonometric ratios.
• Stand the yardstick so it's perpendicular to the ground so that it forms a right angle. The sun will cast a shadow. Mark the end of the shadow with a piece of chalk.
• Measure the length of the shadow. This will be considered the length of the base of the triangle.
Draw a diagram of the triangle made by connecting the top of the yardstick to the marked tip of the shadow. Use inverse trigonometry to determine the angle of elevation. $\tan X = \frac{36 \text{in}}{27 \text{in}}$ $\arctan \frac{36}{27} = X$ $\arctan \frac{4}{3} = X$ $X = 53^\circ$
Step 4) Now, we can use trigonometry to solve the triangle for the height of the bell tower. $\tan 53^\circ = \frac{h}{111 \text{ft}}$ Clearing the fractions, $111 (\tan 53^\circ) = h$ Plugging in the value of $\tan 53^\circ$ gives us $111 \frac{4}{3} = h$ Simplify for $148 \text{ft} = h$ According to our calculations, the height of the Clothier Bell Tower is 148 feet.
## History: Eratosthenes and the Earth
In ancient Greece, mathematician Eratosthenes made a name for himself in the history books by calculating the circumference of the Earth by using shadows. Many other mathematicians had attempted the problem before, but Eratosthenes was the first one to actually have any success. His rate of error was less than 2%.
Eratosthenes used shadows to calculate the distance around the Earth. As an astronomer, he determined the time of the summer solstice when the sun would be directly over the town of Syene in Egypt (now Aswan). On this day, with the sun directly above, there were no shadows, but in Alexandria, which is about 500 miles north of Syene, Eratosthenes saw shadows. He calculated based on the length of the shadow that the angle at which the sun hit the Earth was 7 °. He used this calculation, along with his knowledge of geometry, to determine the circumference of the Earth.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# References
All of the images on this page, unless otherwise stated on their own image page, were made or photographed by the author Richard Scott.
The information on Eratosthenes can be cited to http://www.math.twsu.edu/history/men/eratosthenes.html.
The main image and details about it were found at http://www.imdb.com/title/tt0105378/.
Some of the ideas for problems/pictures on this page are based from ideas or concepts in the Interactive Mathematics Program Textbooks by Fendel, Resek, Alper and Fraser.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://crypto.stackexchange.com/tags/hash-collision/hot
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# Tag Info
## Hot answers tagged hash-collision
40
### Best way to reduce chance of hash collisions: Multiple hashes, or larger hash?
The risk of collision is only theoretical; it will not happen in practice. Time spent worrying about such a risk of collision is time wasted. Consider that even if you have $2^{90}$ 1MB blocks (that's a billion of billions of billions of blocks -- stored on 1TB hard disks, the disks would make a pile as large of the USA and several kilometers high), risks of ...
14
### No SHA-1 Collision? Yet SHA1 is broken?
We call a primitive broken, if there is any attack faster than bruteforce/what we expect of an ideal primitive. Broken does not mean that there are practical attacks. There are no known collisions in SHA-1. Still we call collision resistance of SHA-1 is broken, because there is a theoretical attack that can find collisions using fewer than $2^{80}$ calls to ...
12
### Can I find two specific words with the same md5 hash?
What you want is called a chosen prefix collision. Given p1, p2 you want to find m1, m2 such that hash(p1 || m1) = hash(p2 || m2). Generic attack The generic attack to find this, is creating messages starting with p1 and just as many starting with p2. Thanks to the birthday problem you'll find a match after around 2n/2 messages. For a 128 bit hash like ...
10
### Is SHA-1 still practical secure under specific scenarios?
I would recommend phasing out SHA-1 in any scenario where collision-resistance of a hash is required, for there is a wide consensus that an attack with $2^{69}$ complexity would work, it would already be feasible by a resourceful entity, and attacks only get better. I'm still confident that SHA-1 is preimage and second-preimage resistant for all practical ...
9
### Do MD5's weaknesses affect Oplop?
The short answer is: technically, no. The weaknesses of MD5 are not an issue here. However MD5 is seriously inappropriate, for it is the wrong king of security primitive; also its reputation is tarnished. If a collision attack was to be feared, then using MD5 would be a disaster, for it is now hopelessly broken w.r.t. to collision resistance; but that does ...
8
### Is it fair to assume that SHA1 collisions won't occur on a set of <100k strings
The chance of a collision in such a set is approximately $\frac{1/2 \cdot n^2}{2^{160}}$, which for n=100k evaluates to about $3.4 \cdot 10^{-39}$. So it is fair to say, such a collision won't occur accidentially. AFAIK nobody has every found a SHA-1 collision. Collisions become likely once you generate about $2^{80}$ or $10^{24}$ hashes. If ...
8
### Change in probability of collision when removing digits from MD5 hexadecimal hash values
First things first: there is nothing magical in hexadecimal. The output of MD5 is, nominally, a sequence of 128 bits, or, if you prefer, 16 bytes (each being able to get any value between 0 and 255). Hexadecimal is just a trick to represent a single byte as two characters in a limited range (digits, and letters from 'a' to 'f'). If you can have 30 characters ...
8
### Why would you expect to find a collision in a hash function after approximately $\sqrt{n}$ hashes?
A collision is between two values. If you take a random pair of values you get a 1/2n chance of having a collision. With 2n/2 values you have about 2n-1 pairs, so you could expect about 1/2 chance of collision. (That's just the "intuitive way" of thinking about it; in practice, there are mathematical details.)
8
### Can there be two hash functions without common collisions?
No, in general, there will always be a pair of inputs that will collide for both hash functions. Specifically, if the hash functions have fixed sized outputs, and both take an arbitrary input which is at least as long as the sum of their outputs, then there will be bitstrings $X$ and $Y$ with $X \neq Y$, $A(X) = A(Y)$ and $B(X) = B(Y)$ Here is a simple ...
8
### Increased CRC collision probability when adding bits to input message
First of all, if your goal is to keep the garbled messages to "once every hundred years", well, you already don't meet that goal, even before the change. With an 8 bit CRC, a random change has a probability 1/256 of being accepted; hence if your wireless network has a transmission error at least once every three months (which, to me, sounds like an ...
7
### What is pre-image resistance, and how can the lack thereof be exploited?
Preimage resistance is about the most basic property of a hash function which can be thought. It means: For a given $h$ in the output space of the hash function, it is hard to find any message $x$ with $H(x) = h$. (Note that the it is hard here and in the next definitions is not formally defined, but can be formalized by looking at families of hash ...
7
### Is the number of creatable torrents limited?
1640 is a huge number. For instance, if you consider each torrent to consist in a single byte each (so they are quite uninteresting torrents), and you pack them all on 10 TB hard disks (for a torrent to exist, it must exist on at least one hard disk on the planet), and if each such disk weighs about 100g, then the total weight of the disks is about 24 ...
7
### Question about hash collisions
The general idea is that either one of the inner hashes, or the combining hash must collide, since there is not other place to introduce the collision. Assume we found a collision for H. This means we have X, Y with $X \neq Y$ such that: $h(h(X_0)||h(X_1)) = h(h(Y_0)||h(Y_1))$ Now we define: $A = h(X_0)||h(X_1)$ and $B = h(Y_0)||h(Y_1)$ This gives us a ...
6
### Counter mode secure hash algorithm
First of all, I think I want to correct you at one point; in step 2, you aren't actually that interested in whether the operation is commutative, what you're actually interested in is that the operation is associative, that is, if $(a \oplus b) \oplus c = a \oplus (b \oplus c)$. In essence, your operator $\oplus$ in step 2 turns out to be a group operation. ...
6
### Creating a hash of XOR'd blocks
I believe that this is a poorly written question: such an $h$ obviously doesn't have either preimage resistance, second preimage resistance or collision resistance. The inability to rederive the specific value of $m$ based on its hash is not an interesting property; it's pretty much true of any function which generates an output shorter than its input. I ...
6
### finding collision for truncated SHA1 hash output
echo -n "06b2f82fd81b2c20" | sha1sum e42d65afd2bc126a2e8e609257287084c43fc06a echo -n "02c60cb75083ceef" | sha1sum e42d65afd277988908c01bc539c9d71aff728322 Notice the first ten characters of the SHA1 hash match, indicating a 40-bit match. Other pairs are 0534164decf1166c, 06670357183cba13 and 0addd115537e4b39, 09a3cbdd0d00773b. Note that I am ...
5
### Is calculating a hash code for a large file in parallel less secure than doing it sequentially?
If you want to use Skein (one of the SHA-3 candidates) anyway: it has a "mode of operation" (configuration variant) for tree hashing, which works just like your method 2. It does this internally of the operation, as multiple calls of UBI on the individual blocks. This is described in section 3.5.6 of the Skein specification paper (version 1.3). You will ...
5
### Is calculating a hash code for a large file in parallel less secure than doing it sequentially?
Actually a tree-based hashing as you describe it (your method 2) somewhat lowers resistance to second preimages. For a hash function with a n-bit output, we expect resistance to: collisions up to 2n/2 effort, second preimages up to 2n, preimages up to 2n. "Effort" is here measured in number of invocations of the hash function on a short, "elementary" ...
5
### Are there two known strings which have the same MD5 hash value?
MD5 was intended to be a cryptographic hash function, and one of the useful properties for such a function is its collision-resistance. Ideally, it should take work comparable to around $2^{64}$ tries (as the output size is $128$ bits, i.e. there are $2^{128}$ different possible values) to find a collision (two different inputs hashing to the same output). ...
5
### Using an MD5 hash as a password
Password strength is typically measured in bits of entropy, or in layman's terms, the amount of "true randomness" in the system. This is measured by the process of how the password is generated rather than by the number of bits in the output. It's a simple extension of Kerckhoff's principle: assume your attacker knows your process, and the only information ...
4
### Are there two known strings which have the same MD5 hash value?
A new result shows how to generate single block MD5 collisions, including an example collision: Message 1 Message 2 > md5sum message1.bin message2.bin > 008ee33a9d58b51cfeb425b0959121c9 message1.bin > 008ee33a9d58b51cfeb425b0959121c9 message2.bin There is an earlier example of a single block collision but not technique for generating it was ...
4
### Strength of multiple hash iterations?
In the question's code, the parameter $n=100$ controlling the number of iterations controls (nearly linearly) the cost of evaluating the hash_password function. As a first order approximation, in the attack model where the result and salts are known, the time it takes for an adversary to find the right password in a list by brute force grows (nearly ...
4
### Strength of multiple hash iterations?
Yes. Iterating the hash like you do slightly increases the chance of collisions (as a hash function is not a random permutation, but an approximation of a random function): It is enough if two different passwords produce the same hash at any of the 100 steps, to produce the same final hash. (But as Thomas noted in a comment, the probability of collisions ...
4
### Best way to reduce chance of hash collisions: Multiple hashes, or larger hash?
To have approximately a 50% chance of a collision, you'd need $2^{128}$ data blocks. This comes from the birthday problem. Are you anticipating your list to be that large? I would doubt it as that would be an astronomical amount of data (much, much more than a petabyte). That said, it is very, very unlikely that a collision for MD5 would also be a collision ...
4
### Best way to reduce chance of hash collisions: Multiple hashes, or larger hash?
The risk of collision is only theoretical; it will not happen in practice. Except in one particular instance. The description given implies that this system is going to be some form of de-duplicating filesystem or backup system. For most users, the collision risk is tiny. But, for one particular class of users, there is a much larger risk. Those ...
4
### Are there any known collisions for the SHA-2 family of hash functions?
In short, no. So, what is the current state of cryptanalysis with SHA-1 (for reference only as this question relates to SHA-2) and SHA-2? Bruce Schneier has declared SHA-1 broken. That is because researchers found a way to break full SHA-1 in $2^{69}$ operations. Much less than the $2^{80}$ operations it should take to find a collision due to the birthday ...
4
### How does a birthday attack on a hashing algorithm work?
The method described in the link you cited is based on Floyd's cycle finding algorithm, also known as "the tortoise and the hare" algorithm. This is a general-purpose algorithm for detecting cycles in iterated maps, which I will first describe below. Specifically, consider the sequence $(x_i)$ defined by $x_i = H(x_{i-1})$ for some map $H$ and some initial ...
4
### Do MD5's weaknesses affect Oplop?
Your scheme is a bad one. It is too fast. Also, your step 4 is weird and I worry it might introduce biases into the password that might reduce the entropy of your resulting passwords. Instead, I suggest you use the following alternative: Hash the password using PBKDF2, with the nickname as salt and the iteration count set to make this about as slow as ...
4
### Does a break in a collision resistance property of a hash function by definition implies an attack at the first pre-image attack?
This relation obviously doesn't hold. If you define "break" as faster than what's expected of an ideal hash function Define TrivialCollisionHash = GoodHash(input.Skip(1 bit)). Finding pre-images for this is just as hard as for GoodHash, i.e. $2^n$. Finding a collision is trivial, just flip the first bit. If you define "break" as faster than a certain ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://quant.stackexchange.com/questions/3930/how-to-simulate-cointegrated-prices/3932
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# How to simulate cointegrated prices
Is there any simple way to simulate cointegrated prices?
-
There are some previous questions about correlation rather than cointegration that may be helpful: How to simulate correlated assets for illustrating portfolio diversification? and How to generate a random price series with a specified range and correlation with an actual price? – chrisaycock♦ Aug 9 '12 at 13:52
2
One of the easier ways to think about it is simulating a bivariate VAR system in (ln) levels where there is contemporaneous correlation. So you would begin by simulating the correlated errors (per chrisaycock's link) and then add back in the autoregressive structure. This would give you the simulated log levels, so then just take $exp$ to convert to prices. – John Aug 9 '12 at 15:19
@John Could you please re-post your comment as an answer. – Tal Fishman Aug 9 '12 at 18:28
1
– user508 Aug 9 '12 at 19:13
## 2 Answers
One way to construct cointegrated timeseries it to use the error-correction representation (see Engle, Granger 1987 for details of the equivalence).
To generate two timeseries that are cointegrated, start with your cointegrating vector $(\alpha_1, \alpha_2)$ so that you want $\alpha_1x_t + \alpha_2y_t$ to be stationary; choose initial values $x_0, y_0$ and a parameter $\gamma\in (0,1)$ that controls how strongly cointegrated the series are. Then generate each timestep as:
$x_{t+1} = x_t - \gamma (x_t + (\alpha_2/\alpha_1)y_t) + \epsilon_{1t}$
$y_{t+1} = y_t - \gamma (y_t + (\alpha_1/\alpha_2)x_t) + \epsilon_{2t}$
For price series, it's generally the cumulative returns that you want to be cointegrated. To generate prices, as John mentioned in his comment above, follow the above procedure for log-prices, then exponentiate.
-
Consider a $T \times N$ matrix of potentially cointegrating prices $P$. Define $Y_{t}\equiv ln\left(P_{t}\right)$. In the multivariate framework, there are two basic methods to estimate the cointegrating relationships. The first is an error correction framework of the form $$\Delta Y_{t} = \beta_{0}+\beta_{1}\Delta Y_{t-1}+\beta_{2}Y_{t-1}+\varepsilon_{t}$$ that is most convenient when attempting to perform statistical tests on the coefficients. The alternate approach is a vector autoregressive model of the form $$Y_{t} = \beta_{0}+\beta_{1}Y_{t-1}+\varepsilon_{t}.$$ For the purposes of simulation, they are effectively equivalent. One must estimate $\beta_{0}$ and $\beta_{1}$ and solve for $\varepsilon_{t}$. There are many potential distribution assumptions that one could make about the behavior of $\varepsilon_{t}$, but a simple one would be that it follows a multivariate normal distribution with a mean of zero and a covariance matrix equal to the sample covariance matrix. More complicated assumptions might be that the variances and correlations are time-varying or that there are fat tails. For financial time series, these may be important to consider.
To simulate $\widetilde{Y}_{t+1}$, you thus obtain $\widetilde{\varepsilon}_{t+1}$ by whatever means are appropriate and calculate $$\widetilde{Y}_{t+1} = \beta_{0}+\beta_{1}Y_{t}+\widetilde{\varepsilon}_{t+1}$$
For $i>1$, one would need to be careful to incorporate the simulated values from the previous period so that $$\widetilde{Y}_{t+i} = \beta_{0}+\beta_{1}\widetilde{Y}_{t+i-1}+\widetilde{\varepsilon}_{t+i}$$ in order to ensure the autoregressive features in each simulated path.
After calculating the simulated values of $\widetilde{Y}_{t+i}$, one would want to convert them back to prices by calculating $\widetilde{P}_{t+i}\equiv \mathrm{exp}\left(\widetilde{Y}_{t+i}\right)$.
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http://mathematica.stackexchange.com/questions/15277/plotting-a-complicated-vector-field?answertab=active
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Plotting a complicated vector field
If we consider the vector $\left ( A \cdot \nabla \right) \: B$, we have in Cartesian coordinates
$$\left ( A \cdot \nabla \right) \: B = \left ( A \cdot \nabla B_x \right ) e_x + \left ( A \cdot \nabla B_y \right ) e_y + \left ( A \cdot \nabla B_z \right ) e_z,$$ which gives in full writing:
$$\left ( A \cdot \nabla \right) \: B = \left (A_x \frac{\partial \: B_x}{\partial \: x} + A_y \frac{\partial \: B_x}{\partial \: y} + A_z \frac{\partial \: B_x}{\partial \: z} \right )e_x + \left (A_x \frac{\partial \: B_y}{\partial \: x} + A_y \frac{\partial \: B_y}{\partial \: y} + A_z \frac{\partial \: B_y}{\partial \: z} \right )e_y + \left (A_x \frac{\partial \: B_z}{\partial \: x} + A_y \frac{\partial \: B_z}{\partial \: y} + A_z \frac{\partial \: B_z}{\partial \: z} \right )e_z$$
Now, say $B=A$ and $A=\left (10 \: x, \: 20 \, y^3, \: 30 \:z \right )$.
I would like to plot the field of $A$ and of $\left ( A \cdot \nabla \right) \: A$ in Mathematica. How to go about?
(My objective in this exercise is to know which direction the vector $\left ( A \cdot \nabla \right) \: A$ lies with respect to $A$ and also test for different vectors $A$ when the lines of $\left ( A \cdot \nabla \right) \: A$ are straight or are curved.)
Thanks a lot...
-
next time, please give a try to your question in Mathematica, if only to save time to people answering them having to type the definitions. – chris Nov 27 '12 at 10:18
1 Answer
Let us first define the vector field
````A = {10 x, 20 y^3, 30 z};
````
and load the vector analysis package:
````<< VectorAnalysis`
SetCoordinates[Cartesian[x, y, z]];
````
Now let's define $A\cdot \nabla A$
````field = (A.Grad[#]) & /@ A
(* ==> {100 x, 1200 y^5, 900 z} *)
````
and plot both fields:
````pl1 =
VectorPlot3D[A, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
VectorColorFunction -> "Heat", VectorPoints -> Coarse];
pl2 =
VectorPlot3D[field, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
VectorColorFunction -> "ThermometerColors", VectorPoints -> Coarse];
Show[pl1, pl2]
````
We could also look at a slice
````{VectorPlot[Most[A], {x, 0, 1}, {y, 0, 1},
VectorColorFunction -> "Heat"],
VectorPlot[Most[field], {x, 0, 1}, {y, 0, 1},
VectorColorFunction -> "ThermometerColors"]} // Show
````
Or using a different plotting function, `StreamPlot`:
````pl1 = StreamPlot[Most[A], {x, 0, 1}, {y, 0, 1},
StreamColorFunction -> "SolarColors"];
pl2 = StreamPlot[Most[field], {x, 0, 1}, {y, 0, 1},
StreamColorFunction -> "LakeColors"];
Show[pl1, pl2]
````
Or yet another, `LineIntegralConvolutionPlot` (takes a bit longer)
````pl1 = LineIntegralConvolutionPlot[{Most[A], {"noise", 800, 800}}, {x,
0, 1}, {y, 0, 1}, ColorFunction -> "Heat", LightingAngle -> 0,
LineIntegralConvolutionScale -> 3, Frame -> False];
pl2 = StreamPlot[Most[field], {x, 0, 1}, {y, 0, 1},
StreamColorFunction -> Function[x, White]];
Show[pl1, pl2]
````
-
Who'll be the first to write a version of your answer using the new Grad? :P – Rojo Nov 27 '12 at 10:42
@Rojo doesn't seem to be that different from the old one ;-) – chris Nov 27 '12 at 11:02
lang-mma
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http://unapologetic.wordpress.com/2008/04/29/examples-of-convergent-series/?like=1&source=post_flair&_wpnonce=c1ce40f6e1
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# The Unapologetic Mathematician
## Examples of Convergent Series
Today I want to give two examples of convergent series that turn out to be extremely useful for comparisons.
First we have the geometric series whose terms are the sequence $a_n=a_0r^n$ for some constant ratio $r$. The sequence of partial sums is
$\displaystyle\sum\limits_{k=0}^na_0r^k=a_0\left(1+r+r^2+...+r^n\right)$
If $r\neq1$ we can multiply this sum by $\frac{1-r}{1-r}$ to find
$\displaystyle\sum\limits_{k=0}^na_0r^k=a_0\frac{1-r^{n+1}}{1-r}$
Then as $n$ goes to infinity, this sequence either blows up (for $|r|>1$) or converges to $\frac{a_0}{1-r}$ (for $|r|<1$). In the border case $r=\pm1$ we can also see that the sequence of partial sums fails to converge. Thus the geometric series converges if and only if $|r|<1$, and we have a nice simple formula telling us the sum.
The other one I want to hit is the so-called $p$-series, whose terms are $a_n=n^{-p}$ starting at $n=1$. Here we use the integral test to see that
$\displaystyle\lim\limits_{n\rightarrow\infty}\left(\sum\limits_{k=1}^n\frac{1}{n^p}-\int\limits_1^n\frac{dx}{x^p}\right)=D$
so the sum and integral either converge or diverge together. If $p\neq1$ the integral gives $\frac{n^{1-p}-1}{1-p}$, which converges for $p>1$ and diverges for $p<1$.
If $p=1$ we get $\ln(n)$, which diverges. In this case, though, we have a special name for the limit of the difference $D$. We call it “Euler’s constant”, and denote it by $\gamma$. That is, we can write
$\displaystyle\sum\limits_{k=1}^n\frac{1}{k}=\ln(n)+\gamma+e(n)$
where $e(n)$ is an error term whose magnitude is bounded by $\frac{1}{n}$.
In general we have no good value for the sums of these series, even where they converge. It takes a bit of doing to find $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$, as Euler did in 1735 (solving the “Basel Problem” that had stood for almost a century), and now we have values for other even natural number values of $p$. The sum $\sum\frac{1}{n^3}$ is known as Apéry’s constant, after Roger Apéry who showed that it was irrational in 1979. Yes, we didn’t even know whether it was a rational number or not until 30 years ago. We have basically nothing about odd integer values of $p$.
If we say $s$ instead of $p$, and let $s$ take complex values (no, I haven’t talked about complex numbers yet, but some of you know what they are) we get Riemann’s function $\zeta(s)=\sum\frac{1}{n^s}$, which is connected to some of the deepest outstanding questions in mathematics today.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 4 Comments »
1. [...] if we set , this tells us that . Then the comparison test with the geometric series tells us that [...]
Pingback by | May 5, 2008 | Reply
2. [...] the complex norm is multiplicative, everything for the geometric series goes through again: if , and it diverges if . The case where is more complicated, but it can be [...]
Pingback by | August 28, 2008 | Reply
3. [...] the final summation converges because it’s a geometric series with initial term and ratio . This implies that [...]
Pingback by | December 14, 2009 | Reply
4. [...] this is a chunk of a geometric series; since , the series must converge, and so we can make this sum as small as we please by choosing [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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|
http://mathhelpforum.com/math-topics/115097-2d-plane-intersection-point.html
|
# Thread:
1. ## 2D plane intersection point
Hello all,
i need help with the following question:
Given a 2d plane which has 4 distinct points, C, D, A, B. Points A and B define an infinite line. At time t=0 a player starts moving at a constant velocity from point C to point D, where he arrives at t=1. Determine, if the player hits the wall. If he does, find the corresponding time and position.
I started this problem by taking an infinite length line AB and from the given data , the distance of CD is V(v=velocity from C to D). But I am unable to understand about how to find the intersection of the two lines. I started by trying to find if C and lie on either side of AB or on AB but could not proceed much with this. Can someone plz help me out with problem.
Thanx a lot ,
Aarav
2. Originally Posted by arnav.akash9
Hello all,
i need help with the following question:
Given a 2d plane which has 4 distinct points, C, D, A, B. Points A and B define an infinite line. At time t=0 a player starts moving at a constant velocity from point C to point D, where he arrives at t=1. Determine, if the player hits the wall. If he does, find the corresponding time and position.
Are we to assume that the line AB is the "wall" you mention?
Given points $(x_1, y_1)$ and $(x_2, y_2)$, the equation of the line through them is $(x_2- x_1)(y- y_1)= (y_2- y_1)(x- x_1)$ or $y= \frac{y_2- y_1}{x_2- x_1)(x- x_1)+ y_1$.
Find the equations of the two lines in that form and set the "y"s equal:
The line through A and B is $y= \frac{y_B-y_A}{x_B-x_A}(x- x_A)+ y_A$ and the line through C and D is $y= \frac{y_D-y_C}{x_D-x_C)(x- xC)+ y_C$. Setting those two "y"s equal gives the linear equation $\frac{y_B-y_A}{x_B-x_A}(x- x_A)+ y_A= \frac{y_D-y_C}{x_D-x_C)(x- xC)+ y_C$ which you can solve for x.
Now, go back and write line CD in terms of t. Since the line passes through C at t= 0 and D at t= 1, it can be written as $x= (x_D- x_C)t+ x_C$, $y= (y_D-y_C)t+ y_C$.
Put the value you got for x or y into one of those equations and solve for t.
I started this problem by taking an infinite length line AB and from the given data , the distance of CD is V(v=velocity from C to D). But I am unable to understand about how to find the intersection of the two lines. I started by trying to find if C and lie on either side of AB or on AB but could not proceed much with this. Can someone plz help me out with problem.
Thanx a lot ,
Aarav
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http://math.stackexchange.com/questions/121645/presumably-simple-laplace-transform-question/121647
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# Presumably Simple Laplace Transform Question
I have a (presumably simple) Laplace Transform problem which I'm having trouble with:
$$\mathcal L\big\{t \sinh(4t)\big\} = ?$$
How would I go about solving this? Would you please show working if possible, or alternatively point me in the right direction regarding how to go about solving this?
Also, I'm studying Electrical Engineering at University, and in my ElectEng lectures, my lecturers are referring to Laplace transforms and the Laplace domain (with regard to the frequency response of circuits) in a way which I haven't been exposed to in my mathematics courses (as we've only briefly covered Laplace). Do you know any good resources which I could look at (either on the web or in the library) to get a better understanding of Laplace transforms (and, in particular, their application to circuit analysis)?
Sorry if the last part of this question is out of the scope of this website.
Thanks in advance.
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## 4 Answers
These notes here look good.
The definition of the Laplace transform is $\mathcal{L}(f)(s) := \int_0^\infty e^{-st} f(t) dt$ and for $\sinh$ the following holds: $\sinh x = \frac12 (e^x -e^{-x})$. Now you put these two things together and compute
$$\frac12 \int_0^\infty xe^{4x} e^{-xs} dx - \frac12 \int_0^\infty xe^{-4x} e^{-xs} dx$$
Hope this helps.
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Thanks bro! I don't remember ever begin taught hyperbolic trig functions, but now that I refer to my formula book I see them there so I'll take note. Thanks again! – JonaGik Mar 18 '12 at 9:00
@JonaGik Glad I could help : ) – Matt N. Mar 18 '12 at 9:02
You got $t\operatorname{sinh}{4t}=t\frac{e^{4t}-e^{-4t}}{2}$ , so: $$\mathcal{L}\big\{t\operatorname{sinh}4t\big\}=\tfrac{1}{2}\left(\mathcal{L}\left\{{te^{4t}}\right\}+\mathcal{L}\left\{te^{-4t}\right\}\right)$$
We know that $\mathcal{L}\big\{e^{at}f(t)\big\}=F(s-a)$. Furthermore $\mathcal{L}\left\{t\right\}=\frac{1}{s^2}$. So in our case $\mathcal{L}\left\{te^{4t}\right\}=F(s-4)=\frac{1}{(s-4)^2}$. Finally:$$\mathcal{L}\big\{t \text{sinh} 4t\big\}=\frac{0.5}{(s-4)^2}+\frac{0.5}{(s+4)^2}$$
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Thanks for your help. – JonaGik Mar 18 '12 at 9:22
– chemeng Mar 18 '12 at 10:29
1. There is a generic formula to write $G(s)=\mathcal{L}\{tf(t)\}(s)$ in terms of $F(s)=\mathcal{L}\{f(t)\}(s)$; seeing it involves integration by parts. Have you covered this rule?
2. Hyperbolic sine is a difference of exponentials; can you find the Laplace transform of these?
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Regarding #1, I don't believe I have covered that rule. Could you please point me to it and its derivation, if you don't mind? Regarding #2, that was the main thing I needed to see to solve it. Thanks. – JonaGik Mar 18 '12 at 9:24
@Jona: It's an easy derivation the way Bitrex has it; just note that $te^{-st}=-D_s e^{-st}$ and the derivative can be interchanged with the integral (b/c of sufficient regularity). – anon Mar 18 '12 at 9:53
$\mathcal{L}\{f(t)\}' = F'(s)= \int_0^\infty\frac{d}{ds}[e^{-st} f(t)] dt = \int_0^\infty e^{-st}[-t f(t)]dt$,
so $\mathcal{L}\{tf(t)\} = -F'(s).$
Since $\mathcal{L}\{\sinh(4t)\} = \frac{4}{s^2 - 8}$, (I "cheated" and checked a table!)
Then $\mathcal{L}\{t*\sinh(4t)\} = -\mathcal{L}\{\sinh(4t)\}' = -\frac{d}{ds}(\frac{4}{s^2 - 16})$.
In general, $\mathcal{L}\{t^nf(t)\} = (-1)^n\frac{d}{ds}^nF(s)$.
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http://crypto.stackexchange.com/questions/1858/is-there-some-way-to-generate-a-non-predictable-random-number-in-a-decentralised/3383
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Is there some way to generate a non-predictable random number in a decentralised network?
Is there a way to generate a random number with given restrictions:
• It will be used in a decentralised network with a big number of peers (no central authority to generate it)
• Its generation should not rely on any third-party service (for example, going to a specific website)
• Its generation is triggered by a semi-random data being sent through the network, and it should relate to the data
• The computer sending the data should not have an advantage in determining the random number (for example, computing some number that would have a really high chance of becoming the seeked number)
I was thinking of using a random number like that to counter a 51% attack against the Bitcoin network. The main problem there is that an attacker with a lot of computation power can compute a couple "blocks" in advance and only release them to counter "blocks" generated by legitimate users. I figured a way to counter that would be by requiring generation of some random number that couldn't be pre-computed in advance, but only when a block is sent through the network. Required use of that number would then invalidate blocks that are precomputed beforehand, preventing the attack.
So, is there a way to generate a random number like that described above?
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3 Answers
A service that provides such numbers is called a random beacon.
Since everyone has to agree on what a beacon's value is and peers may not have a complete view of the network, it is very difficult to construct a universally verifiable value using only internal network data. Since data only becomes canonical when it is included in a block (a block that is accepted by everyone else), I'm not sure what else could be used.
If you relax the internal-only constraint and allow external beacons, you will hopefully be able to use NIST's beacon service. You can also construct values with financial data however this only works when markets are open.
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The classic way to do this is to have all parties commit to individual random values by publishing a secure hash of a suitably random-nonce-padded number. Once the commitments have been distributed, the parties open the commitments by publishing the nonce and the number. The numbers are combined in some previously agreed suitable fashion such as adding them modulo some number or xoring them together. As long as your value was included then you are confident that the result is random. If the numbers used are large enough, the nonce is not required.
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This is a well-known problem from the secure multi-party computation literature, known as the coin-tossing problem. Several people want to get together and jointly generate an unbiased coin toss, where the security property is that no one can influence the bias of the coin.
The problem is impossible if adversaries are allowed to be computationally unbounded. The simplest protocol in the computationally bounded setting is the following (I'll describe it for generating a single bit -- it generalizes naturally for generating multiple bits):
1. Party $i$ publishes a commitment to a secret randomly chosen bit $b_i$.
2. After everyone has posted their commitment, everyone opens their commitments. The final coin is taken to be $b_1 \oplus \cdots \oplus b_n$.
As long as one person generates their $b_i$ honestly, the outcome is an unbiased coin. This follows from the hiding and binding properties of the commitment scheme.
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Actually, your protocol is vulnerable to a classic attack. Suppose there are two parties, Alice and Bob, and Bob goes last. Then, Bob can ensure the final bit will be 0 as follows. Bob can copy Alice's commitment, then copy Alice's opening of the commitment, which ensures that $b_1=b_2$ and thus the final coin is always 0. To fix this, you could compute the final coin as $H(b_1,\dots,b_n)$, or use other standard fixes. – D.W. Sep 7 '12 at 8:24
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http://mathoverflow.net/questions/14858/k3-surface-of-genus-8
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## K3 surface of genus 8
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $V$ be a complex vector space of dimension 6 and let $G\subset {\mathbb P}^{14}\simeq {\mathbb P}(\Lambda^2V)$ be the image of the Plucker embedding of the Grassmannian $Gr(2, V)$.
1. Why the degree of $G$ is 14? or in general, how to calculate the degree of a Plucker embedding?
Let ${\mathbb P}^8\simeq L\subset {\mathbb P}(\Lambda^2V)$ be a generic 8-plane and $S$ be the intersection of $L$ with $G$.
1. How to prove that $S$ is a K3 surface?
Another question: the paper said that this construction depends on 19 parameters. I know that this is the dimension of the deformation family of the polarized K3 we get here. But I think that in this statement 19 is coming from varying the generic 8-plane. How can we obtain this number?
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## 3 Answers
To be able calculate the degree it is worth to read a bit of Griffiths-Harris about Grassmanians (chapter 1 section 5). To prove that $S$ is $K3$ one needs to caluclate the canonical bundle of $G$, use simple facts about Plucker embedding, use adjunction formula and finally the fact that a simply connected surface with $K\cong O$ is $K3$.
I will make the second bit of calculation, that proves that $S$ is a $K3$ (so I don't calculate the degree $14$).
First we want to calculate the canonical bundle of $G$. Denote by $E$ the trivial $6$-dimensional bundle over $G$, and by $S$ the universal (tautological) rank $2$ sub-bundle. Then the tangent bundle to $G$ is $TG=S^* \otimes (E/S)$ . It follows from the properties of $c_1$ that $c_1(TG)=6c_1(S^*)$. Similarly for the canonical bundle $K_G$ we have the expression $K_G\cong (detS^*)^{\otimes -6}$.
Now we will use the (simple) statement from Griffiths-Harris that under the Plucker embedding we have the isomorphism of the line bundles $det S^*=O(1)$. Using the previous calculation we see $K_G\cong O(-6)$. Finally, the surface $S$ is an iterated (6 times) hyperplane section of $G$. So by Lefshetz theorem it has same fundamental group as $G$, i.e., it is simply-connected. It suffices now to see that its canonical bundle is trivial. This is done using the adjunction formula $K_D=K_X+D|_D$. Every time we cut $G$ by a hyper-plane we tensor the canonical by $O(1)$, but $O(-6)\otimes O(6)\cong O$.
Added. The calculation of the degree is done by Andrea
Added. The number 19 is obtaied in the following way. The dimension of the grassmanian of $8$-planes in $CP^{14}$ is $6\cdot 9=54$. At the same time the Grassmanian of $2$-planes in $\mathbb C^6$ has symmetires, given by $SL(6,\mathbb C)$, whose dimension is 35. We should quotient by these symmetries and get $54-35=19$
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Many thanks to both of you. – Guangbo Xu Feb 10 2010 at 16:09
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Guess you are reading Beauville-Donagi :-)
To compute the degree of the Grassmannian you can note that the hyperplane divisor under the Plucker embedding has class $\sigma_1$, so the degree of the Grassmannian is just $\sigma_1^8$, which you can compute by Schubert calculus. See the paragraph about Grassmannians in Griffiths-Harris.
Namely $\sigma_1^2 = \sigma_{1,1} + \sigma_2$ hence $\sigma_1^3 = 2\sigma_{2,1} + \sigma_3$ by Pieri's formula and finally $\sigma_1^4 = \sigma_4 + 3 \sigma_{3,1} + 2 \sigma_{2,2}$ again by Pieri.
Since $\sigma_4$, $\sigma_{3,1}$ and $\sigma_{2,2}$ are Poincaré dual to themselves, you find $\sigma_1^8 = 1 + 3^2 + 2^2 = 14$
The other question has already been answered by Dmitri.
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Yes, Beauville-Donagi:) – Guangbo Xu Feb 10 2010 at 16:09
1) In general, the degree of the Grassmannian $G(k,n)$ in the Plucker embedding is given by $$(k(n-k))!\prod_{i=1}^k\frac{(i-1)!}{(n-k+i-1)!}$$This is calculated by finding $\sigma_1^{k(n-k)}$ using Pieri's rule. See this link for more details.
See Dimitri's answer for question 2).
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http://math.stackexchange.com/questions/146844/how-to-divide-aleph-numbers?answertab=oldest
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# How to divide aleph numbers
Recently, I was wondering how division of aleph numbers would work. First, I thought about how finite cardinality division would work. What I came up with was that the result of $A/B$ where $A$ and $B$ are both cardinalities, is the number of times that each element of $B$ had to be mapped to an element of $A$ in order to ensure that all elements of $A$ were mapped to.
Extending this to aleph numbers, specifically, I thought that $\aleph_1/\aleph_0$ would be $\aleph_1$. The reasoning behind this is that there are an infinite ($\aleph_1$) number of real numbers between any two natural numbers. As such, that mapping would need to be applied.
Can anyone validate this idea? Is this how division of aleph numbers actually works, or am I totally off base?
Thanks.
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Yes, $\aleph_1/\aleph_0 = \aleph_1$ is the only thing that makes sense. But what about $\aleph_0/\aleph_0$ ... shouldn't you do that one first? And what about $\aleph_0/\aleph_1$ ... – GEdgar May 18 '12 at 22:02
## 2 Answers
Much like I wrote in Cardinal number subtraction, if $\kappa$ and $\lambda$ are two $\aleph$-numbers, it might be possible to define division, but this definition would have to be limited and awkward.
If $\kappa$ and $\lambda$ are both regular cardinals and $\kappa<\lambda$ then every partition of $\lambda$ into $\kappa$ many parts would have to have at least one part would be of size $\lambda$. In a sense this means that $\frac\lambda\kappa=\lambda$. This is indeed the case with $\aleph_1/\aleph_0$, both are regular cardinals are $\aleph_0<\aleph_1$.
If, however, $\kappa=\lambda$ this is no longer defined, since $\kappa=2\cdot\kappa=\aleph_0\cdot\kappa=\ldots=\kappa\cdot\kappa=\ldots$, so there can be many partitions of $\lambda$ into $\kappa$ many parts, and in each the parts would vary in size (singletons; pairs; countably infinite sets; etc.)
When $\lambda$ is a singular limit cardinal, e.g. $\aleph_\omega$ this breaks down completely, since singular cardinals can be partitioned into a "few" "small" parts. In the $\aleph_\omega$ case these would be parts of size $\aleph_n$ for every $n$, which make a countable partition in which all parts are smaller than $\aleph_\omega$.
The only reasonable way I can think that cardinal division can be defined would have to consider the Surreal numbers, and the embedding of the ordinals in them. However this will not be compatible with cardinal arithmetic at all (the surreal numbers form a field).
I should also remark that your reasoning for $\aleph_1/\aleph_0$ being $\aleph_1$ is invalid. First note that neither is a real number, and that it is possible that $\aleph_1$ is much smaller than the cardinality of the real numbers (so between two natural numbers there are a lot more real numbers). Secondly, note that between two rational numbers there are also infinitely many rational numbers - does that mean $\aleph_0/\aleph_0=\aleph_0$?
However your rationale is not that far off, as I remarked in the top part of the post, if you take a set of size $\aleph_1$ and partition it into $\aleph_0$ many parts you are guaranteed that at least one of the parts would have size $\aleph_1$.
Further reading:
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Very nice. Do we need some sort of choice principle to determine that, given a partition of $\aleph_1$ into $\aleph_0$-many parts, at least one of the parts must have size $\aleph_1$? I understood that it is consistent with ZF that $\aleph_0$ is the only regular cardinal, so doesn't that mean that, for example, $\aleph_1$ can be realized as a countable union of countable subsets? – Cameron Buie May 18 '12 at 23:25
@Cameron: Of course. I assume choice through and through in this answer. Indeed it is consistent with ZF (and without large cardinals at all) that $\aleph_1$ is a countable union of countable sets. If you want two successive singular cardinals you need some pretty large cardinals in the background, though. – Asaf Karagila May 18 '12 at 23:39
If you're looking for something compatible with cardinal multiplication, you'll have to deal with the fundamental problem that $\kappa\cdot\lambda=\max\{\kappa,\lambda\}$ whenever $\kappa,\lambda$ are well-orderable cardinals and at least one of them is an aleph. That sort of absorption means--for example--that $\aleph_0\cdot\aleph_2=\aleph_1\cdot\aleph_2=\aleph_2\cdot\aleph_2=\aleph_2$, so even trying to define $\aleph_2/\aleph_2$ in some way compatible with cardinal multiplication is problematic. Now, one could choose a convention for $\lambda/\kappa$ in instances that $\kappa\leq\lambda$--say, for example, that it's always just $\lambda$--but trying to compatibly define it when $\kappa>\lambda$ is fruitless.
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What is the difference between being well-orderable and being aleph? I was under the impression that the aleph numbers are defined as well-orderable cardinalities. – Dustan Levenstein May 19 '12 at 2:14
Apologies for the ambiguity. The definition of aleph that I learned was infinite well-orderable cardinalities (i.e: $\aleph_0$ is the least aleph). Does that clarify things sufficiently? – Cameron Buie May 19 '12 at 3:17
oh, yes, of course. I wasn't even thinking about finite cardinals. – Dustan Levenstein May 19 '12 at 12:18
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http://math.stackexchange.com/questions/203347/cutting-the-corners-of-a-cube?answertab=active
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# Cutting the corners of a cube
Do you know of any way to cut the corners of a cube by means of rotation assuming that the cube is centered in the origin of XoYZ?
For example if we have a square centered in the origin of XoY and we rotate this square 45 degrees the rotated square will cut all the corners of the original one with equal cuts. I do not know how to generelize this for higher dimensions.
Thanks, Bogdan.
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Well, I guess rotating the cube 45 degrees along the x axis, 45 degress along the z axis, and 45 degrees along the x=y, z=0 axis should do the job. – roman Sep 27 '12 at 9:02
## 2 Answers
A square has $2\cdot2$ sides and $2^2$ vertices, but a three-dimensional cube $C$ has $2\cdot3$ faces and $2^3$ vertices; so you can never cut off all vertices at the same time, using a rotated copy of $C$.
You can, however, do the following: When $C=[{-1},1]^3$ then consider for a suitable $\epsilon>0$ the octahedron $$O:=\bigl\{(x,y,z)\ \bigm|\ |x|+|y|+|z|\leq 3-\epsilon\bigr\}\ .\qquad(*)$$ This will cut off the eight vertices of $C$ allright. In addition, this idea can be generalized to $n$ dimensions, $n\geq2$.
That $(*)$ describes an octahedron is seen as follows: The vertex $(1,1,1)$ in the first octant is cut off in a symmetric way with respect to the three axes by means of the condition $x+y+z\leq 3-\epsilon$. Writing $|x|+|y|+|z|\leq3-\epsilon$ instead of $x+y+z\leq 3-\epsilon$ makes the resulting solid symmetric with respect to all three reflections $x\mapsto -x$, $y\mapsto -y$, $z\mapsto -z$. In $n$ dimensions the corresponding condition would read as $|x_1|+|x_2|+\ldots+|x_n|\leq n-\epsilon$.
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The generalization can be:|x1|+|x2|+ ... |xn|≤n−ϵ ? – Bogdan Sep 27 '12 at 10:35
@Bogdan: See my edit. – Christian Blatter Sep 27 '12 at 11:11
Thank you for your answer. I am looking at the dual polyhedron from the linear programming theory point of view. The cube is a set of inequalities 0<=x<=1, 0<=y<=1, 0<=z<=1. This is something like A * [x, y, z]' <= b. I would like to know if the dual polyhedron can also be expressed in a similar form. – Bogdan Sep 27 '12 at 11:19
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You can cut the corners of a cube by another concentric and congruent cube: taking it in random orientation, it will in most cases not contain any of the corners, thereby "cutting them all off". But the corners will never be cut off regularly in this case for a simple reason: the cutting is done by the faces of a cube of which there are $6$, but there are $8$ corners to cut, so necessarily some distinct corners will be cut off by the same face (which will cut off the whole edge between them as well).
You can cut off the corners of a cube regularly using an octahedron though. In general you need a dual polytope; the dual of a square happens to be a square. The same happens for any regular $n$-gon in the plane, and for the tetrahedron in space, but not for any other (regular) solids.
@Bogdan: You cannot obtain equations for the dual polyhedron directly from those of the polyhedron, as again their numbers do not match ($6$ for the cube, $8$ for the octahedron). The equations for the dual polyhedron come from the vertices of the original polyhedron, and vice versa. – Marc van Leeuwen Sep 27 '12 at 11:19
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http://quant.stackexchange.com/questions/2579/law-of-an-integrated-cir-process-as-sum-of-independent-random-variables
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# Law of an integrated CIR Process as sum of Independent Random Variables
It is known (see for example Joshi-Chan "Fast and Accureate Long Stepping Simulation of the Heston SV Model" available at SSRN) that for a CIR process defined as :
$$dY_t= \kappa(\theta -Y_t)dt+ \omega \sqrt{Y_t}dW_t$$ $Y_0=Y$ together with a correct set of constraints on parameters' value.
Then the law of $\int_0^T Y_t dt$ conditionaly on $Y_0,Y_T$ can be seen as the sum of three (rather complicated) independent random variables (see proposition 4 eq 2.10 in Joshi Chan article)
NB : The original result is coming from Glassermann and Kim "Gamma Expansion of the Heston Stochastic Volatility Model" available at SSRN, but I'm more used to Joshi Chan's expression.
So here is my question :
Does the integrated CIR process itself by any chance has such a representation in the form of the sum of independent random variables ?
PS: The Laplace transform has a known closed-form expression but I couldn't infer directly from this such a representation.
Edit : As Tal has opened a bounty on this here is the Laplace transform of the integrated CIR process :
$$\mathcal{L}\left\{\int_0^t Y_s ds\right\}(\lambda)=\mathbb{E}\left[e^{-\lambda\int_0^t Y_s ds}\right]=e^{-A_\lambda(t)-Y_0.G_\lambda(t)}$$ with $A_\lambda(t)=-\frac{2\kappa.\theta }{\omega^2}. \mathrm{Ln}\left[\frac{2\gamma.e^{(\gamma+\kappa).s/2}}{\gamma.(e^{s.\gamma}+1)+\kappa.(e^{s.\gamma}-1)}\right]$ and $G_\lambda(t)=\frac{2.\lambda.(e^{s.\gamma}-1)}{\gamma.(e^{s.\gamma}+1)+\kappa.(e^{s.\gamma}-1)}$ where $\gamma=\sqrt{\kappa^2+\omega^2.\lambda}$
This is coming from Chesnay, Jeanblanc-Picqué, Yor "Mathematical Methods for Financial Markets" Proposition 6.3.4.1
Best regards
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http://mathhelpforum.com/calculus/107009-equation-normal-curve-parallel-given-line.html
|
# Thread:
1. ## The equation of the normal to a curve which is parallel to a given line
this is the question:
find the equation of the normal to the curve y=3+2x-x^2 which is parallel to the line 2y-x-3=0
2. Originally Posted by mastermin346
this is the question:
find the equation of the normal to the curve y=3+2x-x^2 which is parallel to the line 2y-x-3=0
The gradient of 2y-x-3=0 is 1/2. Therefore the gradient of the required normal is 1/2 therefore the gradient of the tangent is -2. So solve dy/dx = -2 to get the x-coordinate and hence y-coordinate of the required point on the curve. Now you have a point and you have a gradient and therefore you can write down the equation of the normal line.
3. ## WRITE it
Originally Posted by mr fantastic
The gradient of 2y-x-3=0 is 1/2. Therefore the gradient of the required normal is 1/2 therefore the gradient of the tangent is -2. So solve dy/dx = -2 to get the x-coordinate and hence y-coordinate of the required point on the curve. Now you have a point and you have a gradient and therefore you can write down the equation of the normal line.
can u write it?it dificult to me.
4. Originally Posted by mastermin346
this is the question:
find the equation of the normal to the curve y=3+2x-x^2 which is parallel to the line 2y-x-3=0
i am sure you know that the gradient of $y=1/2x+3/2$ is 1/2
$y=-x^2+2x+3$
$\frac{dy}{dx}=-2x+2$ ... gradient of tangent
so gradient of normal will be $-\frac{1}{2-2x}$
have it equal to 1/2 . Do you know why ?
then solve for x , sub into the original equation to get y.
After that , make use the formula that i gave you in my post tp your other question to get the equation .
5. Originally Posted by mastermin346
can u write it?it dificult to me.
You have spent no more than 8 minutes thinking about what I posted before aksing for more help. You need to spend more time than that thinking about the question and the reply I gave.
Have you tried doing what I suggested? What don't you understand? Where do you get stuck?
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|
http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn/30741
|
# How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$
How can if find the sum for:
$$\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$$
I know the answer thanks to Wolfram Alpha. I'm more concerned with how to get to to that number. It cites tests to prove that it is convergent, but my class has never learned these before so I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
-
## 8 Answers
No need to use Taylor Series, this can be derived in a similar way to the formula for geometric series. Lets find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$
Notice that $$S_{m}-rS_{m}=-mr^{m+1}+\sum_{n=1}^{m}r^{n}$$
$$=-mr^{m+1}+\frac{r-r^{m+1}}{1-r}=\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}.$$ Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum. That is $$\sum_{n=1}^{\infty}\frac{2n}{3^{n+1}}=\frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}}$$
$$=\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}}=\frac{1}{2}.$$
Added note:
We can define $$S_m^k(r) =\sum_{n=1}^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.
This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_k^m(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli Numbers. In particular, the denominator is $(1-r)^{k+1}$.
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That's a great answer, much simpler than how I've approached this question before. – mixedmath♦ May 27 '11 at 23:09
As indicated in other answers, you can reduce this to summing $\displaystyle{\sum_{n=1}^\infty na^n}$ with $|a|<1$ (by pulling out the constant $\frac{2}{3}$ and rewriting with $a=\frac{1}{3}$). This in turn can be reduced to summing geometric series by rearranging and factoring. Note that, assuming everything converges nicely (which it does):
$\begin{matrix} &a & + & 2a^2 & + & 3a^3 &+& 4a^4 &+& \cdots\\ =&a &+& a^2 &+& a^3 &+& a^4 &+& \cdots\\ +& & & a^2 &+& a^3 &+& a^4 &+& \cdots\\ +& & & & & a^3 &+& a^4 &+& \cdots\\ +& & & & & & & a^4 &+& \cdots\\ +& & & & & & & & & \vdots \end{matrix}$
Factoring out the lowest power of $a$ in each row yields
$\begin{align*} \sum_{n=1}^\infty na^n &= a(1+a^2+a^3+\cdots)\\ &+ a^2(1+a^2+a^3+\cdots)\\ &+ a^3(1+a^2+a^3+\cdots)\\ &+ a^4(1+a^2+a^3+\cdots)\\ &\vdots \end{align*}$
Each row in the last expression has the common factor $a(1+a+a^2+a^3+\cdots)$, and factoring this out yields
$\begin{align*}\sum_{n=1}^\infty na^n &=a(1+a+a^2+a^3+\cdots)(1+a+a^2+a^3+\cdots)\\ &=a(1+a+a^2+a^3+\cdots)^2.\end{align*}$
Now you can finish by summing the geometric series.
Eric Naslund's answer was posted while I was writing, but I thought that this alternative approach might be worth posting. I also want to mention that in general one should be careful about rearranging series as though they were finite sums. To be more formal, some of the steps above would require justification based on absolute convergence.
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If you want a solution that doesn't require derivatives or integrals, notice that \begin{eqnarray} 1+2x+3x^2+4x^3+\dots = 1 + x + x^2 + x^3 + \dots \\ + x + x^2+ x^3 + \dots\\ + x^2 + x^3 + \dots \\ +x^3 + \dots \\ + \dots \\ =1 + x + x^2 + x^3+\dots \\ +x(1+x+x^2+\dots) \\ +x^2(1+x+\dots)\\ +x^3(1+\dots)\\ +\dots \\ =(1+x+x^2+x^3+\dots)^2=\frac{1}{(1-x)^2} \end{eqnarray}
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I was going to say that! You are fast! – dot dot Oct 29 '12 at 22:44
wow this is dam clever – Justin Meltzer Oct 29 '12 at 22:47
Factor out the 2/3. Then write $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=2}^{\infty} \frac{1}{3^n} + \sum_{n=3}^{\infty} \frac{1}{3^n} + \cdots$$
It is easy to show that $$\sum_{n=m}^{\infty} \frac{1}{3^n} = \frac{3}{2} \left(\frac{1}{3} \right)^m$$ and so $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{2} \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n$$ which you can sum. Don't forget to put the 2/3 back in.
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Hints
1. You know (don't you?) the formula for $S(a) = \sum_{n=0}^\infty a^n$ for $|a| < 1$
2. Take the derivative (with respect to $a$) of both sides to obtain a formula for $\sum_{n=1}^\infty n a^n$
3. Show that your series can be put in that form.
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Thanks for answering. 1) No I don't know that formula 2) Can you explain by what you mean by derive both sides? – Backus Apr 3 '11 at 21:59
1
– leonbloy Apr 3 '11 at 22:05
Yeah I don't know calculus – Backus Apr 3 '11 at 22:06
The definition of 'convergence' is based on calculus. So without knowing calculus, it is pretty hard to understand anything about series. Perhaps you can tell us why you are interested in this particular sum? – wildildildlife Apr 3 '11 at 22:32
@wildildildlife I'm in a class that covers precalculus and calculus in one year, and we are transitioning currently. For whatever reason we learned series early. Our class still understood convergence without calculus though. – Backus Apr 3 '11 at 23:32
Note that $\int \left( 1 + 2x + 3x^2 + 4x^3 + \ldots \right)dx = x + x^2 + x^3 + \ldots + c$, i.e., a geometric series, which converges to $x/(1 - x)$. So $$\frac{d}{dx} \left( \frac{x}{1 - x} \right) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2}.$$ Therefore $$1 + 2x + 3x^2 + 4x^3 + \ldots = \frac{1}{(1 - x)^2}.$$
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+1, but wouldn't it have been simpler to say that it was the derivative of $1+x+x^2+...$, converging to $\frac1{1-x}$? – Mike Oct 29 '12 at 22:46
I decided to start with what was given, so it is easier for the OP to see. – glebovg Oct 29 '12 at 22:48
Justin, observe that this only holds for $|x| < 1$, otherwise your sum simplifies to $\infty$. – glebovg Oct 29 '12 at 22:54
What I meant is that you chose $c=0$ while $c=1$ is a more well known series and is easier to take the derivative of. – Mike Oct 29 '12 at 23:23
You can find by differentiation. Just notice that $(x^n)' = nx^{n-1}$. By the theory of power series we obtain (by uniform convergence on any compact subset of $(-1,1)$) that $$\left(\sum_{n=1}^\infty x^n\right)' = \sum_{n=1}^\infty (x^n)' = \sum_{n=1}^\infty n x^{n-1}.$$ The sum on the left hand side is equal to $\left(\frac{x}{1-x}\right)'$. You need to notice that your sum can be written in a similar way as $\sum_{n=1}^\infty nx^{n-1}$.
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Thank you for helping, but I have never learned differentiation. – Backus Apr 3 '11 at 21:58
My favorite proof of this is here
I also have the following method for $\sum_{n=1}^\infty {n\over 2^{n-1}}$ (one can use a similar method for $\sum_{n=1}^\infty {n\over3^n}$):
We first show that $\sum\limits_{n=7}^\infty {n\over 2^{n-1}} ={1\over4}$.
We start with a rectangle of width 1 and height $1/4$. Divide this into eights:
Now divide each eighth-rectangle above in half and take 7 of them. This gives $A_1={7\over 2^6}$.
There are $2\cdot8-7=9$ boxes left over, each having area $2^{-6}$.
Divide each remaining $16^{\rm th}$-rectangle in half and take 8 of them. This gives $A_2={7\over 2^6}+{8\over 2^7}$.
There are $2\cdot9-8=10$ boxes left over, each having area $2^{-7}$.
Divide each remaining $32^{\rm nd}$-rectangle in half and take 9 of them. This gives $A_3={7\over 2^6}+{8\over 2^7}+{9\over 2^8}$.
There are $2\cdot10-9=11$ boxes left over, each having area $2^{-8}$.
Divide each remaining $64^{\rm th}$-rectangle in half and take 10 of them. This gives $A_4={7\over 2^6}+{8\over 2^7}+{9\over 2^8}+{10\over2^9}$.
There are $2\cdot11-9=12$ boxes left over, each having area $2^{-9}$.
At each stage, we double the number of remaining boxes, keeping the same leftover area, and take approximately half of them to form the next term of the series.
At the $n^{\rm th}$ stage, we have $$A_n= {7\over 2^6}+{8\over 2^7}+\cdots+{6+n\over2^{5+n}},$$
with leftover area $$2(n+7)-(n+6)\over 2^{n+5}.$$
It follows that, $${7\over2^6}+{8\over2^7}+{9\over2^8}+\cdots= {1\over4}.$$ Consequently, $$\sum_{n=1}^\infty{n\over 2^{n-1}}= \sum_{n=1}^6 {n\over 2^{n-1}} +\sum_{n=7}^\infty{n\over 2^{n-1}} ={15\over 4}+{1\over4}=4.$$
You can also "Fubini" this (I think this is what Jonas is doing).
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http://www.sagemath.org/doc/reference/modules/sage/modules/fg_pid/fgp_module.html
|
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• Modules »
# Finitely generated modules over a PID¶
You can use Sage to compute with finitely generated modules (FGM’s) over a principal ideal domain R presented as a quotient V/W, where V and W are free.
NOTE: Currently this is only enabled over R=ZZ, since it has not been tested and debugged over more general PIDs. All algorithms make sense whenever there is a Hermite form implementation. In theory the obstruction to extending the implementation is only that one has to decide how elements print. If you’re annoyed that by this, fix things and post a patch!
We represent M=V/W as a pair (V,W) with W contained in V, and we internally represent elements of M non-canonically as elements x of V. We also fix independent generators g[i] for M in V, and when we print out elements of V we print their coordinates with respect to the g[i]; over $$\ZZ$$ this is canonical, since each coefficient is reduce modulo the additive order of g[i]. To obtain the vector in V corresponding to x in M, use x.lift().
Morphisms between finitely generated R modules are well supported. You create a homomorphism by simply giving the images of generators of M0 in M1. Given a morphism phi:M0–>M1, you can compute the image of phi, the kernel of phi, and using y=phi.lift(x) you can lift an elements x in M1 to an element y in M0, if such a y exists.
TECHNICAL NOTE: For efficiency, we introduce a notion of optimized representation for quotient modules. The optimized representation of M=V/W is the quotient V’/W’ where V’ has as basis lifts of the generators g[i] for M. We internally store a morphism from M0=V0/W0 to M1=V1/W1 by giving a morphism from the optimized representation V0’ of M0 to V1 that sends W0 into W1.
The following TUTORIAL illustrates several of the above points.
First we create free modules V0 and W0 and the quotient module M0. Notice that everything works fine even though V0 and W0 are not contained inside $$\ZZ^n$$, which is extremely convenient.
```sage: V0 = span([[1/2,0,0],[3/2,2,1],[0,0,1]],ZZ); W0 = V0.span([V0.0+2*V0.1, 9*V0.0+2*V0.1, 4*V0.2])
sage: M0 = V0/W0; M0
Finitely generated module V/W over Integer Ring with invariants (4, 16)
```
The invariants are computed using the Smith normal form algorithm, and determine the structure of this finitely generated module.
You can get the V and W used in constructing the quotient module using V() and W() methods:
```sage: M0.V()
Free module of degree 3 and rank 3 over Integer Ring
Echelon basis matrix:
[1/2 0 0]
[ 0 2 0]
[ 0 0 1]
sage: M0.W()
Free module of degree 3 and rank 3 over Integer Ring
Echelon basis matrix:
[1/2 4 0]
[ 0 32 0]
[ 0 0 4]
```
We note that the optimized representation of M0, mentioned above in the technical note has a V that need not be equal to V0, in general.
```sage: M0.optimized()[0].V()
Free module of degree 3 and rank 2 over Integer Ring
User basis matrix:
[0 0 1]
[0 2 0]
```
Create elements of M0 either by coercing in elements of V0, getting generators, or coercing in a list or tuple or coercing in 0. Coercing in a list or tuple takes the corresponding linear combination of the generators of M0.
```sage: M0(V0.0)
(0, 14)
sage: M0(V0.0 + W0.0) # no difference modulo W0
(0, 14)
sage: M0([3,20])
(3, 4)
sage: 3*M0.0 + 20*M0.1
(3, 4)
```
We make an element of M0 by taking a difference of two generators, and lift it. We also illustrate making an element from a list, which coerces to V0, then take the equivalence class modulo W0.
```sage: x = M0.0 - M0.1; x
(1, 15)
sage: x.lift()
(0, -2, 1)
sage: M0(vector([1/2,0,0]))
(0, 14)
sage: x.additive_order()
16
```
Similarly, we construct V1 and W1, and the quotient M1, in a completely different 2-dimensional ambient space.
```sage: V1 = span([[1/2,0],[3/2,2]],ZZ); W1 = V1.span([2*V1.0, 3*V1.1])
sage: M1 = V1/W1; M1
Finitely generated module V/W over Integer Ring with invariants (6)
```
We create the homomorphism from M0 to M1 that sends both generators of M0 to 3 times the generator of M1. This is well defined since 3 times the generator has order 2.
```sage: f = M0.hom([3*M1.0, 3*M1.0]); f
Morphism from module over Integer Ring with invariants (4, 16) to module with invariants (6,) that sends the generators to [(3), (3)]
```
We evaluate the homomorphism on our element x of the domain, and on the first generator of the domain. We also evaluate at an element of V0, which is coerced into M0.
```sage: f(x)
(0)
sage: f(M0.0)
(3)
sage: f(V0.1)
(3)
```
Here we illustrate lifting an element of the image of f, i.e., finding an element of M0 that maps to a given element of M1:
```sage: y = f.lift(3*M1.0); y
(0, 13)
sage: f(y)
(3)
```
We compute the kernel of f, i.e., the submodule of elements of M0 that map to 0. Note that the kernel is not explicitly represented as a submodule, but as another quotient V/W where V is contained in V0. You can explicitly coerce elements of the kernel into M0 though.
```sage: K = f.kernel(); K
Finitely generated module V/W over Integer Ring with invariants (2, 16)
sage: M0(K.0)
(2, 0)
sage: M0(K.1)
(3, 1)
sage: f(M0(K.0))
(0)
sage: f(M0(K.1))
(0)
```
We compute the image of f.
```sage: f.image()
Finitely generated module V/W over Integer Ring with invariants (2)
```
Notice how the elements of the image are written as (0) and (1), despite the image being naturally a submodule of M1, which has elements (0), (1), (2), (3), (4), (5). However, below we coerce the element (1) of the image into the codomain, and get (3):
```sage: list(f.image())
[(0), (1)]
sage: list(M1)
[(0), (1), (2), (3), (4), (5)]
sage: x = f.image().0; x
(1)
sage: M1(x)
(3)
```
TESTS:
```sage: from sage.modules.fg_pid.fgp_module import FGP_Module
sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ)
sage: W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = FGP_Module(V, W); Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: Q([1,3])
(1, 3)
sage: Q(V([1,3,4]))
(0, 11)
sage: Q(W([1,16,0]))
(0, 0)
sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],QQ)
sage: W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1])
sage: Q = FGP_Module(V, W); Q
Finitely generated module V/W over Rational Field with invariants (0)
sage: q = Q.an_element(); q
(1)
sage: q*(1/2)
(1/2)
sage: (1/2)*q
(1/2)
```
AUTHOR:
• William Stein, 2009
sage.modules.fg_pid.fgp_module.FGP_Module(V, W, check=True)¶
INPUT:
• V – a free R-module
• W – a free R-submodule of V
• check – bool (default: True); if True, more checks on correctness are performed; in particular, we check the data types of V and W, and that W is a submodule of V with the same base ring.
OUTPUT:
• the quotient V/W as a finitely generated R-module
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: import sage.modules.fg_pid.fgp_module
sage: Q = sage.modules.fg_pid.fgp_module.FGP_Module(V, W)
sage: type(Q)
<class 'sage.modules.fg_pid.fgp_module.FGP_Module_class_with_category'>
sage: Q is sage.modules.fg_pid.fgp_module.FGP_Module(V, W, check=False)
True
```
class sage.modules.fg_pid.fgp_module.FGP_Module_class(V, W, check=True)¶
Bases: sage.modules.module.Module
A finitely generated module over a PID presented as a quotient V/W.
INPUT:
• V – an R-module
• W – an R-submodule of V
• check – bool (default: True)
EXAMPLES:
```sage: A = (ZZ^1)/span([[100]], ZZ); A
Finitely generated module V/W over Integer Ring with invariants (100)
sage: A.V()
Ambient free module of rank 1 over the principal ideal domain Integer Ring
sage: A.W()
Free module of degree 1 and rank 1 over Integer Ring
Echelon basis matrix:
[100]
sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: type(Q)
<class 'sage.modules.fg_pid.fgp_module.FGP_Module_class_with_category'>
```
Element¶
alias of FGP_Element
Hom(N)¶
EXAMPLES:
```sage: V = span([[1/2,0,0],[3/2,2,1],[0,0,1]],ZZ); W = V.span([V.0+2*V.1, 9*V.0+2*V.1, 4*V.2])
sage: Q = V/W
sage: Q.Hom(Q)
Set of Morphisms from Finitely generated module V/W over Integer Ring with invariants (4, 16) to Finitely generated module V/W over Integer Ring with invariants (4, 16) in Category of modules over Integer Ring
sage: M = V/V.zero_submodule()
sage: M.Hom(Q)
Set of Morphisms from Finitely generated module V/W over Integer Ring with invariants (0, 0, 0) to Finitely generated module V/W over Integer Ring with invariants (4, 16) in Category of modules over Integer Ring
```
V()¶
If this module was constructed as a quotient V/W, returns V.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.V()
Free module of degree 3 and rank 3 over Integer Ring
Echelon basis matrix:
[1/2 0 0]
[ 0 1 0]
[ 0 0 1]
```
W()¶
If this module was constructed as a quotient V/W, returns W.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.W()
Free module of degree 3 and rank 3 over Integer Ring
Echelon basis matrix:
[1/2 8 0]
[ 0 12 0]
[ 0 0 4]
```
annihilator()¶
Return the ideal of the base ring that annihilates self. This is precisely the ideal generated by the LCM of the invariants of self if self is finite, and is 0 otherwise.
EXAMPLES:
```sage: V = span([[1/2,0,0],[3/2,2,1],[0,0,1]],ZZ); W = V.span([V.0+2*V.1, 9*V.0+2*V.1, 4*V.2])
sage: Q = V/W; Q.annihilator()
Principal ideal (16) of Integer Ring
sage: Q.annihilator().gen()
16
sage: Q = V/V.span([V.0]); Q
Finitely generated module V/W over Integer Ring with invariants (0, 0)
sage: Q.annihilator()
Principal ideal (0) of Integer Ring
```
base_ring()¶
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.base_ring()
Integer Ring
```
cardinality()¶
Return the cardinality of this module as a set.
EXAMPLES:
```sage: V = ZZ^2; W = V.span([[1,2],[3,4]]); A = V/W; A
Finitely generated module V/W over Integer Ring with invariants (2)
sage: A.cardinality()
2
sage: V = ZZ^2; W = V.span([[1,2]]); A = V/W; A
Finitely generated module V/W over Integer Ring with invariants (0)
sage: A.cardinality()
+Infinity
sage: V = QQ^2; W = V.span([[1,2]]); A = V/W; A
Vector space quotient V/W of dimension 1 over Rational Field where
V: Vector space of dimension 2 over Rational Field
W: Vector space of degree 2 and dimension 1 over Rational Field
Basis matrix:
[1 2]
sage: A.cardinality()
+Infinity
```
coordinate_vector(x, reduce=False)¶
Return coordinates of x with respect to the optimized representation of self.
INPUT:
• x – element of self
• reduce – (default: False); if True, reduce coefficients modulo invariants; this is ignored if the base ring isn’t ZZ.
OUTPUT:
The coordinates as a vector. That is, the same type as self.V(), but in general with fewer entries.
EXAMPLES:
```sage: V = span([[1/4,0,0],[3/4,4,2],[0,0,2]],ZZ); W = V.span([4*V.0+12*V.1])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 0, 0)
sage: Q.coordinate_vector(-Q.0)
(-1, 0, 0)
sage: Q.coordinate_vector(-Q.0, reduce=True)
(3, 0, 0)
```
If x isn’t in self, it is coerced in:
```sage: Q.coordinate_vector(V.0)
(1, 0, -3)
sage: Q.coordinate_vector(Q(V.0))
(1, 0, -3)
```
TESTS:
```sage: V = span([[1/2,0,0],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: Q.coordinate_vector(Q.0 - Q.1)
(1, -1)
sage: O, X = Q.optimized()
sage: O.V()
Free module of degree 3 and rank 2 over Integer Ring
User basis matrix:
[0 0 1]
[0 2 0]
sage: phi = Q.hom([Q.0, 4*Q.1])
sage: x = Q(V.0); x
(0, 4)
sage: Q.coordinate_vector(x, reduce=True)
(0, 4)
sage: Q.coordinate_vector(-x, reduce=False)
(0, -4)
sage: x == 4*Q.1
True
sage: x = Q(V.1); x
(0, 1)
sage: Q.coordinate_vector(x)
(0, 1)
sage: x == Q.1
True
sage: x = Q(V.2); x
(1, 0)
sage: Q.coordinate_vector(x)
(1, 0)
sage: x == Q.0
True
```
cover()¶
If this module was constructed as V/W, returns the cover module V. This is the same as self.V().
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.V()
Free module of degree 3 and rank 3 over Integer Ring
Echelon basis matrix:
[1/2 0 0]
[ 0 1 0]
[ 0 0 1]
```
gen(i)¶
Return the i-th generator of self.
INPUT:
• i – integer
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: Q.gen(0)
(1, 0)
sage: Q.gen(1)
(0, 1)
sage: Q.gen(2)
Traceback (most recent call last):
...
ValueError: Generator 2 not defined
sage: Q.gen(-1)
Traceback (most recent call last):
...
ValueError: Generator -1 not defined
```
gens()¶
Returns tuple of elements $$g_0,...,g_n$$ of self such that the module generated by the gi is isomorphic to the direct sum of R/ei*R, where ei are the invariants of self and R is the base ring.
Note that these are not generally uniquely determined, and depending on how Smith normal form is implemented for the base ring, they may not even be deterministic.
This can safely be overridden in all derived classes.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.gens()
((1, 0), (0, 1))
sage: Q.0
(1, 0)
```
has_canonical_map_to(A)¶
Return True if self has a canonical map to A, relative to the given presentation of A. This means that A is a finitely generated quotient module, self.V() is a submodule of A.V() and self.W() is a submodule of A.W(), i.e., that there is a natural map induced by inclusion of the V’s. Note that we do not require that this natural map be injective; for this use is_submodule().
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: A = Q.submodule((Q.0, Q.0 + 3*Q.1)); A
Finitely generated module V/W over Integer Ring with invariants (4, 4)
sage: A.has_canonical_map_to(Q)
True
sage: Q.has_canonical_map_to(A)
False
```
hom(im_gens, codomain=None, check=True)¶
Homomorphism defined by giving the images of self.gens() in some fixed fg R-module.
Note
We do not assume that the generators given by self.gens() are the same as the Smith form generators, since this may not be true for a general derived class.
INPUTS:
• im_gens – a list of the images of self.gens() in some R-module
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: phi = Q.hom([3*Q.1, Q.0])
sage: phi
Morphism from module over Integer Ring with invariants (4, 12) to module with invariants (4, 12) that sends the generators to [(0, 3), (1, 0)]
sage: phi(Q.0)
(0, 3)
sage: phi(Q.1)
(1, 0)
sage: Q.0 == phi(Q.1)
True
```
This example illustrates creating a morphism to a free module. The free module is turned into an FGP module (i.e., quotient V/W with W=0), and the morphism is constructed:
```sage: V = span([[1/2,0,0],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (2, 0, 0)
sage: phi = Q.hom([0,V.0,V.1]); phi
Morphism from module over Integer Ring with invariants (2, 0, 0) to module with invariants (0, 0, 0) that sends the generators to [(0, 0, 0), (1, 0, 0), (0, 1, 0)]
sage: phi.domain()
Finitely generated module V/W over Integer Ring with invariants (2, 0, 0)
sage: phi.codomain()
Finitely generated module V/W over Integer Ring with invariants (0, 0, 0)
sage: phi(Q.0)
(0, 0, 0)
sage: phi(Q.1)
(1, 0, 0)
sage: phi(Q.2) == V.1
True
```
Constructing two zero maps from the zero module:
```sage: A = (ZZ^2)/(ZZ^2); A
Finitely generated module V/W over Integer Ring with invariants ()
sage: A.hom([])
Morphism from module over Integer Ring with invariants () to module with invariants () that sends the generators to []
sage: A.hom([]).codomain() is A
True
sage: B = (ZZ^3)/(ZZ^3)
sage: A.hom([],codomain=B)
Morphism from module over Integer Ring with invariants () to module with invariants () that sends the generators to []
sage: phi = A.hom([],codomain=B); phi
Morphism from module over Integer Ring with invariants () to module with invariants () that sends the generators to []
sage: phi(A(0))
()
sage: phi(A(0)) == B(0)
True
```
A degenerate case:
```sage: A = (ZZ^2)/(ZZ^2)
sage: phi = A.hom([]); phi
Morphism from module over Integer Ring with invariants () to module with invariants () that sends the generators to []
sage: phi(A(0))
()
```
The code checks that the morphism is valid. In the example below we try to send a generator of order 2 to an element of order 14:
```sage: V = span([[1/14,3/14],[0,1/2]],ZZ); W = ZZ^2
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (2, 14)
sage: Q([1,11]).additive_order()
14
sage: f = Q.hom([Q([1,11]), Q([1,3])]); f
Traceback (most recent call last):
...
ValueError: phi must send optimized submodule of M.W() into N.W()
```
invariants(include_ones=False)¶
Return the diagonal entries of the smith form of the relative matrix that defines self (see _relative_matrix()) padded with zeros, excluding 1’s by default. Thus if v is the list of integers returned, then self is abstractly isomorphic to the product of cyclic groups $$Z/nZ$$ where $$n$$ is in $$v$$.
INPUT:
• include_ones – bool (default: False); if True, also include 1’s in the output list.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.invariants()
(4, 12)
```
An example with 1 and 0 rows:
```sage: V = ZZ^3; W = V.span([[1,2,0],[0,1,0], [0,2,0]]); Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (0)
sage: Q.invariants()
(0,)
sage: Q.invariants(include_ones=True)
(1, 1, 0)
```
is_finite()¶
Return True if self is finite and False otherwise.
EXAMPLES:
```sage: V = span([[1/2,0,0],[3/2,2,1],[0,0,1]],ZZ); W = V.span([V.0+2*V.1, 9*V.0+2*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 16)
sage: Q.is_finite()
True
sage: Q = V/V.zero_submodule(); Q
Finitely generated module V/W over Integer Ring with invariants (0, 0, 0)
sage: Q.is_finite()
False
```
is_submodule(A)¶
Return True if self is a submodule of A. More precisely, this returns True if if self.V() is a submodule of A.V(), with self.W() equal to A.W().
Compare has_canonical_map_to().
EXAMPLES:
```sage: V = ZZ^2; W = V.span([[1,2]]); W2 = W.scale(2)
sage: A = V/W; B = W/W2
sage: B.is_submodule(A)
False
sage: A = V/W2; B = W/W2
sage: B.is_submodule(A)
True
```
This example illustrates that this command works in a subtle cases.:
```sage: A = ZZ^1
sage: Q3 = A / A.span([[3]])
sage: Q6 = A / A.span([[6]])
sage: Q6.is_submodule(Q3)
False
sage: Q6.has_canonical_map_to(Q3)
True
sage: Q = A.span([[2]]) / A.span([[6]])
sage: Q.is_submodule(Q6)
True
```
linear_combination_of_smith_form_gens(x)¶
Compute a linear combination of the optimised generators of this module as returned by smith_form_gens().
EXAMPLE:
```sage: X = ZZ**2 / span([[3,0],[0,2]], ZZ)
sage: X.linear_combination_of_smith_form_gens([1])
(1)
```
ngens()¶
Return the number of generators of self.
(Note for developers: This is just the length of gens(), rather than of the minimal set of generators as returned by smith_form_gens(); these are the same in the FGP_Module_class, but not necessarily in derived classes.)
EXAMPLES:
```sage: A = (ZZ**2) / span([[4,0],[0,3]], ZZ)
sage: A.ngens()
1
```
This works (but please don’t do it in production code!)
```sage: A.gens = lambda: [1,2,"Barcelona!"]
sage: A.ngens()
3
```
optimized()¶
Return a module isomorphic to this one, but with V replaced by a submodule of V such that the generators of self all lift trivially to generators of V. Replace W by the intersection of V and W. This has the advantage that V has small dimension and any homomorphism from self trivially extends to a homomorphism from V.
OUTPUT:
• Q – an optimized quotient V0/W0 with V0 a submodule of V such that phi: V0/W0 –> V/W is an isomorphism
• Z – matrix such that if x is in self.V() and c gives the coordinates of x in terms of the basis for self.V(), then c*Z is in V0 and c*Z maps to x via phi above.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: O, X = Q.optimized(); O
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: O.V()
Free module of degree 3 and rank 2 over Integer Ring
User basis matrix:
[0 0 1]
[0 1 0]
sage: O.W()
Free module of degree 3 and rank 2 over Integer Ring
Echelon basis matrix:
[ 0 12 0]
[ 0 0 4]
sage: X
[0 4 0]
[0 1 0]
[0 0 1]
sage: OV = O.V()
sage: Q(OV([0,-8,0])) == V.0
True
sage: Q(OV([0,1,0])) == V.1
True
sage: Q(OV([0,0,1])) == V.2
True
```
random_element(*args, **kwds)¶
Create a random element of self=V/W, by creating a random element of V and reducing it modulo W.
All arguments are passed onto the random_element method of V.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.random_element()
(1, 10)
```
relations()¶
If this module was constructed as V/W, returns the relations module V. This is the same as self.W().
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.relations()
Free module of degree 3 and rank 3 over Integer Ring
Echelon basis matrix:
[1/2 8 0]
[ 0 12 0]
[ 0 0 4]
```
smith_form_gen(i)¶
Return the i-th generator of self. A private name (so we can freely override gen() in derived classes).
INPUT:
• i – integer
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: Q.smith_form_gen(0)
(1, 0)
sage: Q.smith_form_gen(1)
(0, 1)
```
smith_form_gens()¶
Return a set of generators for self which are in Smith normal form.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W
sage: Q.smith_form_gens()
((1, 0), (0, 1))
sage: [x.lift() for x in Q.smith_form_gens()]
[(0, 0, 1), (0, 1, 0)]
```
submodule(x)¶
Return the submodule defined by x.
INPUT:
• x – list, tuple, or FGP module
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2])
sage: Q = V/W; Q
Finitely generated module V/W over Integer Ring with invariants (4, 12)
sage: Q.gens()
((1, 0), (0, 1))
```
We create submodules generated by a list or tuple of elements:
```sage: Q.submodule([Q.0])
Finitely generated module V/W over Integer Ring with invariants (4)
sage: Q.submodule([Q.1])
Finitely generated module V/W over Integer Ring with invariants (12)
sage: Q.submodule((Q.0, Q.0 + 3*Q.1))
Finitely generated module V/W over Integer Ring with invariants (4, 4)
```
A submodule defined by a submodule:
```sage: A = Q.submodule((Q.0, Q.0 + 3*Q.1)); A
Finitely generated module V/W over Integer Ring with invariants (4, 4)
sage: Q.submodule(A)
Finitely generated module V/W over Integer Ring with invariants (4, 4)
```
Inclusion is checked:
```sage: A.submodule(Q)
Traceback (most recent call last):
...
ValueError: x.V() must be contained in self's V.
```
sage.modules.fg_pid.fgp_module.is_FGP_Module(x)¶
Return true of x is an FGP module, i.e., a finitely generated module over a PID represented as a quotient of finitely generated free modules over a PID.
EXAMPLES:
```sage: V = span([[1/2,1,1],[3/2,2,1],[0,0,1]],ZZ); W = V.span([2*V.0+4*V.1, 9*V.0+12*V.1, 4*V.2]); Q = V/W
sage: sage.modules.fg_pid.fgp_module.is_FGP_Module(V)
False
sage: sage.modules.fg_pid.fgp_module.is_FGP_Module(Q)
True
```
sage.modules.fg_pid.fgp_module.random_fgp_module(n, R=Integer Ring, finite=False)¶
Return a random FGP module inside a rank n free module over R.
INPUT:
• n – nonnegative integer
• R – base ring (default: ZZ)
• finite – bool (default: True); if True, make the random module finite.
EXAMPLES:
```sage: import sage.modules.fg_pid.fgp_module as fgp
sage: fgp.random_fgp_module(4)
Finitely generated module V/W over Integer Ring with invariants (4)
```
sage.modules.fg_pid.fgp_module.random_fgp_morphism_0(*args, **kwds)¶
Construct a random fgp module using random_fgp_module, then construct a random morphism that sends each generator to a random multiple of itself. Inputs are the same as to random_fgp_module.
EXAMPLES:
```sage: import sage.modules.fg_pid.fgp_module as fgp
sage: fgp.random_fgp_morphism_0(4)
Morphism from module over Integer Ring with invariants (4,) to module with invariants (4,) that sends the generators to [(0)]
```
sage.modules.fg_pid.fgp_module.test_morphism_0(*args, **kwds)¶
EXAMPLES:
```sage: import sage.modules.fg_pid.fgp_module as fgp
sage: s = 0 # we set a seed so results clearly and easily reproducible across runs.
sage: set_random_seed(s); v = [fgp.test_morphism_0(1) for _ in range(30)]
sage: set_random_seed(s); v = [fgp.test_morphism_0(2) for _ in range(30)]
sage: set_random_seed(s); v = [fgp.test_morphism_0(3) for _ in range(10)]
sage: set_random_seed(s); v = [fgp.test_morphism_0(i) for i in range(1,20)]
sage: set_random_seed(s); v = [fgp.test_morphism_0(4) for _ in range(50)] # long time
```
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http://physics.stackexchange.com/questions/12120/calculating-time-for-a-fully-charged-ups/12121
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# Calculating time for a fully charged UPS
I have a UPS of 1000 Volts connected with 2 batteries each of 150 Amp. How much time it will take to consume the whole UPS (after fully charged) when a device of 1Amp is getting electricity form that UPS.
Please also explain me the calculation.
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## 3 Answers
There are some missing data in your question.
• What voltage does the batteries have, I'm going to assume 12V since it is common.
• Battery capacity, you typed it as 150A but I guess it was 150Ah. Please note that a normal car battery on a car like a VW Golf has approx 60Ah so 150Ah is a quite big battery.
• Output power, you state that you have 1000V (Volts) out, but I guess you are talking about 1000W (Watts). And that should be maximum power.
• Output voltage should be either 230V (or 110V). I will assume 230V.
The first thing is that a device connected to the UPS and is using 1A, that is 1A at 230V or $230V * 1A = 230W.$
Then we go inside the UPS and have a look. We still has the 230W but since it is now a 12V system, we need to reverse the conversion into amps again and get something like $\frac{230W}{12V}=19A$.
And then we divide the battery capacity with this current and get something like, $\frac{(150*2)Ah}{19A}=15h$.
Please note that those 15 hours is the best value you can expect to see, but since there is losses in the conversion from 12V to 230V we probably loose some 10-20% of the energy, and that translate directly into a shorter time.
Let's say 80% of 15h would be approx 12h.
Then I must add that Fortunato does have a point. Batteries degrade over time and can't hold the same charge, so make sure you have some margin and check/service your UPS on yearly basis.
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Unless they are special deep discharge batteries (ie just cheap car/alarm lead-acid) you can typically only take 50% of the rated capacity without damaging them. – Martin Beckett Jul 11 '11 at 3:45
Battery capacity is normally rated in ampere-hours (Ah), not ampere. This means what it sounds like - if you have 2 * 150 Ah of batteries, you can (ideally) pull 1A for 300 hours like in your question (*), or 300 amps for 1 hour. This is an intuitive value of battery capacity but is not strictly speaking a measurement of the energy stored since this depends on the voltage which you pull the current at (and both of these vary during the discharge of a real battery).
You wrote that your batteries are rated at 150A, but you might have misread the 150 Ah figure, in which case your answer is above. You could also have quoted the maximum current capacity of the battery of 150A (typical of a car battery for example), which isn't related to the capacity at all and I agree with the other answer here that you cannot calculate it without further information.
To use the old water analogy - the voltage is, roughly speaking, the water pressure of the battery and the ampere is the rate of water in the stream coming from it. You would have no indication of the size of the reservoir so you cannot calculate for how long your device can run.
(*) This is of course assuming that you can connect your device at all, i.e. an UPS with an output of 1000 volts will not work very well if you connect it to a domestic 110V or 220V piece of equipment :)
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Don't think you can calculate this. I depends on the type, age and history of the batteries. All his factors make a big difference. If you rely need to know the TCT "total charge time" put them threw the cycle once and measure it. And remember this will change with time.
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## protected by Qmechanic♦Apr 9 at 7:56
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://mathhelpforum.com/advanced-math-topics/160033-ratio-currents.html
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# Thread:
1. ## ratio of currents
I solved this first in that this appears to be an inversion.
$\frac{\frac{V^2}{R_2}}{\frac{V^2}{R_1}}<br /> \Rightarrow<br /> \frac{I_1}{I_2}<br /> \Rightarrow<br /> \frac{[1+60(.005)]}{[1+60(.004)]}<br /> \times<br /> \frac{[1+10(.004)]}{[1+10(.005)]}<br /> \Rightarrow<br /> \frac{(1.3)(1.04)}{(1.24)(1.05)} = 1.0384<br />$
I still don't understand what these temperature coefficients mean. or why the have the $C^{-1}$
appreciate any comments...
2. Here's how I read it:
the resistance of a resistor is being modeled as
$R=R_{0}(1+T\alpha),$ where $T$ is temperature in degrees C, and alpha is what is called the "temperature coefficient of resistance". $R_{0}$ is some initial resistance. It would, technically, be the resistance when the temperature is $0^{\circ}\text{C}.$ The resistance of a resistor changes with temperature, and here is one model for how it changes. The units of $\alpha$ must be inverse degrees C in order to cancel out the units of $T$. Otherwise, you wouldn't have units of Ohms on both sides. Hence the $C^{-1}.$ It would, perhaps, be better to write, for example, $\alpha_{1}=0.004(^{\circ}\text{C})^{-1}.$
Does this explanation clear things up?
3. Originally Posted by bigwave
I solved this first in that this appears to be an inversion.
$\frac{\frac{V^2}{R_2}}{\frac{V^2}{R_1}}<br /> \Rightarrow<br /> \frac{I_1}{I_2}<br /> \Rightarrow<br /> \frac{[1+60(.005)]}{[1+60(.004)]}<br /> \times<br /> \frac{[1+10(.004)]}{[1+10(.005)]}<br /> \Rightarrow<br /> \frac{(1.3)(1.04)}{(1.24)(1.05)} = 1.0384<br />$
I still don't understand what these temperature coefficients mean. or why the have the $C^{-1}$
appreciate any comments...
The resistors are in parallel so the potential drop across them is the same, in which case the current ratio is the reciprocal of the power ratio.
CB
4. thanks for the help, I see how this works now, I am finding EE a very interesting application of math.
was wondering if MHF considered an EE catagory... probabaly not enough potential subscribers tho.
5. You might be surprised how many people on MHF can do at least the basics of EE. Certainly digital logic and sophomore-level circuits are well within the capabilities of folks here. Also, signal analysis with Z transforms, Laplace transforms, etc., are all covered here. Maybe not antenna design!
Keep posting questions. I'm finding them interesting, at least. The worst thing that could happen would be that no one knows how to help!
6. ## will definitly keep posting the EE problems
I will definitly keep posting the EE problems.
Its a brave new world for me
I did some study with electronics about 20 years ago. but the math was not the focus.
just building kits and stuff..
now I don't see how it could even be understood at all without math
regards.
r
7. Ok, have a good one.
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http://mathhelpforum.com/trigonometry/81517-rotation-triangle-print.html
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# Rotation of triangle
Printable View
• March 30th 2009, 01:25 PM
Ian1779
1 Attachment(s)
Rotation of triangle
Hi
I am finding this question tough (attached), I think I may have stumbled on the answer but could do with some confirmation I'm on the right path.
Right, so far I have found the length $BC = \sqrt 82$ using Phythagoras.
Now I need to determine $sin \theta$, $cos \theta$ and $tan \theta$
This is where I am stuck. It is not a right angled triangle so can I use SOHCAHTOA?
Never the less I have attempted this way and got
$tan \theta = 1/9$
$sin \theta = 1/\sqrt82$
$cos \theta = 9/\sqrt82$
Which all give the same answer incidentally - is this the right approach or am I off in the completely wrong direction?
I get the last part about the two line notation, but am not sure what values to use for x and y.
Thanks
• March 31st 2009, 12:18 PM
earboth
1 Attachment(s)
Quote:
Originally Posted by Ian1779
Hi
I am finding this question tough (attached), I think I may have stumbled on the answer but could do with some confirmation I'm on the right path.
Right, so far I have found the length $BC = \sqrt 82$ using Phythagoras.
Now I need to determine $sin \theta$, $cos \theta$ and $tan \theta$
This is where I am stuck. It is not a right angled triangle so can I use SOHCAHTOA?
Never the less I have attempted this way and got
$tan \theta = 1/9$
$sin \theta = 1/\sqrt82$
$cos \theta = 9/\sqrt82$
Which all give the same answer incidentally - is this the right approach or am I off in the completely wrong direction?
I get the last part about the two line notation, but am not sure what values to use for x and y.
Thanks
After the rotation the side B''C'' has the slope m = 0. Calculate the the slope of BC (2-point-formula of a straight line) which is equal of the tangens of the angle of rotation:
$m=\dfrac{2-1}{-4-5}=-\dfrac19$
That means you have to rotate the triangle by an angles whose tangens is $+\dfrac19$
I'm not quite sure what is meant by the two-line notation. The angle in question is included by the lines with the equations:
$l_1: y = 0\text{ and }l_2:y = -\dfrac19 \cdot x$
• March 31st 2009, 01:20 PM
Ian1779
Thanks so much. The drawing helped even more so I could visualise it better.
All times are GMT -8. The time now is 09:12 AM.
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http://mathhelpforum.com/algebra/146135-optimization-problems-print.html
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# Optimization Problems
Printable View
• May 23rd 2010, 01:36 PM
lorica2000
Optimization Problems
I need an example or more of optimation problems - finding maximum values.
Thanks
• May 23rd 2010, 03:37 PM
SpringFan25
wow, choose your own question :D
I'll make an easy one then.
Find the maximum of $y=-(x-1)^2 + 15$
Firstly, this is a familiar graph (upside down parabola, shifted around). The maximum value of y will be 15, occuring at x=1.
Now, to prove it:
Step 1: Find the derivative of the function you want to maximuse
$f'(x)=-2(x-1) +0$
Step 2: Set the derivative to zero to find x:
$0=-2(x-1)$
$x=1$
Step 3: Find the value of y that corresponds to this value of x
$y=-(x-1)^2 + 15$
$y=-0^2 + 15$
$y= 15$
Step 4:Check you have a maximum
You can do this by checking the y values on either side are less than 15, or by checking that the second derivative is negative
Explanation
We found the maximum by looking for the point where the gradient was 0. This is the point where the curve stops increasing/decreasing, so it will be a maximum or a minimum.
All times are GMT -8. The time now is 10:05 AM.
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http://en.m.wikipedia.org/wiki/Unit_hyperbola
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# Unit hyperbola
The Unit Hyperbola is blue, its conjugate is green, and the asymptotes are red.
In geometry, the unit hyperbola is the set of points (x,y) in the Cartesian plane that satisfies $x^2 - y^2 = 1 .$ In the study of indefinite orthogonal groups, the unit hyperbola forms the basis for an alternative radial length
$r = \sqrt {x^2 - y^2} .$
Whereas the unit circle surrounds its center, the unit hyperbola requires the conjugate hyperbola $y^2 - x^2 = 1$ to complement it in the plane. This pair of hyperbolas share the asymptotes y = x and y = −x. When the conjugate of the unit hyperbola is in use, the alternative radial length is $r = \sqrt {y^2 - x^2} .$
The unit hyperbola finds applications where the circle must be replaced with the hyperbola for purposes of analytic geometry. A prominent instance is the depiction of spacetime as a pseudo-Euclidean space. There the asymptotes of the unit hyperbola form a light cone. Further, the attention to areas of hyperbolic sectors by Gregoire de Saint-Vincent led to the logarithm function and the modern parametrization of the hyperbola by sector areas. When the notions of conjugate hyperbolas and hyperbolic angles are understood, then the classical complex numbers, which are built around the unit circle, can be replaced with numbers built around the unit hyperbola.
## Asymptotes
Main article: asymptote
Generally asymptotic lines to a curve are said to converge toward the curve. In algebraic geometry and the theory of algebraic curves there is a different approach to asymptotes. The curve is first interpreted in the projective plane using homogeneous coordinates. Then the asymptotes are lines that are tangent to the projective curve at a point at infinity, thus circumventing any need for a distance concept and convergence. In a common framework (x, y, z) are homogeneous coordinates with the line at infinity determined by the equation z = 0. For instance, C. G. Gibson wrote:[1]
For the standard rectangular hyperbola $f = x^2 - y^2 -1$ in R2 the corresponding projective curve is $F = x^2 - y^2 - z^2,$ which meets z = 0 at the points P = (1 : 1 : 0) and Q = (1 : −1 : 0). Both P, Q are simple on F, with tangents x + y = 0, x − y = 0; thus we recover the familiar 'asymptotes' of elementary geometry.
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## Minkowski diagram
Main article: Minkowski diagram
The Minkowski diagram is drawn in a spacetime plane where the spatial aspect has been restricted to a single dimension. The units of distance and time on such a plane are
• units of 30 centimetres length and nanoseconds, or
• astronomical units and intervals of 8 minutes and 20 seconds, or
• light years and years.
Each of these scales of coordinates results in photon connections of events along diagonal lines of slope plus or minus one. Five elements constitute the diagram Herman Minkowski used to describe the relativity transformations: the unit hyperbola, its conjugate hyperbola, the axes of the hyperbola, a diameter of the unit hyperbola, and the conjugate diameter. The plane with the axes refers to a resting frame of reference. The diameter of the unit hyperbola represents a frame of reference in motion with rapidity a where $\tanh \ a = y/x$ and (x,y) is the endpoint of the diameter on the unit hyperbola. The conjugate diameter represents the spatial hyperplane of simultaneity corresponding to rapidity a. In this context the unit hyperbola is a calibration hyperbola[2][3]
Commonly in relativity study the hyperbola with vertical axis is taken as primary:
The arrow of time goes from the bottom to top of the figure — a convention adopted by Richard Feynman in his famous dagrams. Space is represented by planes perpendicular to the time axis. The here and now is a singularity in the middle.[4]
The vertical time axis convention stems from Minkowski in 1908, and is also illustrated on page 48 of Eddington's The Nature of the Physical World (1928).
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## Parametrization
Main article: hyperbolic angle
As a particular conic, the hyperbola can be parametrized by the process of addition of points on a conic. The following description is given by Prasolov & Solovyev (1997):
Fix a point E on the conic. Consider the points at which the straight line drawn through E parallel to AB intersects the conic a second time to be the sum of the points A and B.
For the hyperbola $x^2 - y^2 = 1$ with the fixed point E = (1,0) the sum of the points $(x_1,\ y_1)$ and $(x_2,\ y_2)$ is the point $(x_1 x_2 + y_1 y_2,\ y_ 1 x_2 + y_2 x_1 )$ under the parametrization $x = \cosh \ t$ and $y = \sinh \ t$ this addition corresponds to the addition of the parameter t .
The branches of the unit hyperbola evolve as the points (cosh α, sinh α) and (−cosh α, −sinh α) depending on the hyperbolic angle parameter α
This parameter is hyperbolic angle, which is the argument of the hyperbolic functions.
One finds an early expression of the parametrized unit hyperbola in Elements of Dynamic (1878) by W. K. Clifford. He describes quasi-harmonic motion in a hyperbola as follows:
The motion $\rho = \alpha \cosh(nt + \epsilon) + \beta \sinh(nt + \epsilon)$ has some curious analogies to elliptic harmonic motion. ... The acceleration $\ddot{\rho} = n^2 \rho \ ;$ thus it is always proportional to the distance from the centre, as in elliptic harmonic motion, but directed away from the centre.(pages 89, 90)
A direct way to parameterizing the unit hyperbola starts with the hyperbola xy = 1 parameterized with the exponential function: $( e^t, \ e^{-t}).$
This hyperbola is transformed into the unit hyperbola by a linear mapping having the matrix $A = \tfrac {1}{2}\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}\ :$
$(e^t, \ e^{-t}) \ A = (\frac{e^t + e^{-t}}{2},\ \frac{e^t - e^{-t}}{2}) = (\cosh t,\ \sinh t).$
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## Complex plane algebra
Main article: split-complex number
Whereas the unit circle is associated with complex numbers, the unit hyperbola is key to the split-complex number plane consisting of z = x + y j where j 2 = +1. Then jz = y + x j so that the action of j on the plane is to swap the coordinates. In particular, this action swaps the unit hyperbola with its conjugate, and also swaps pairs of conjugate diameters of the hyperbola
In terms of the hyperbolic angle parameter a, the unit hyperbola consists of points
$\pm(\cosh a + j \sinh a)$ where j = (0,1).
The right branch of the unit hyperbola corresponds to the positive coefficient. In fact, this branch is the image of the exponential map acting on the j-axis. Since
$\exp(aj) exp(bj) = \exp((a+b)j),$
the branch is a group under multiplication. Unlike the circle group, this unit hyperbola group is not compact. Similar to the ordinary complex plane, a point, not on the diagonals, has a polar decomposition using the parametrization of the unit hyperbola and the alternative radial length.
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## References
1. C.G. Gibson (1998) Elementary Geometry of Algebraic Curves, p 159, Cambridge University Press ISBN 0-521-64140-3
2. Anthony French (1968) Special Relativity, page 83, W. W. Norton & Company
3. W.G.V. Rosser (1964) Introduction to the Theory of Relativity, figure 6.4, page 256, London: Butterworths
4. A.P. French (1989) "Learning from the past; Looking to the future", acceptance speech for 1989 Oersted Medal, American Journal of Physics 57(7):587–92
• William Kingdon Clifford (1878) Elements of Dynamic, pages 89 & 90, London: MacMillan & Co; on-line presentation by Cornell University Historical Mathematical Monographs.
• F. Reese Harvey (1990) Spinors and calibrations, Figure 4.33, page 70, Academic Press, ISBN 0-12-329650-1 .
• Viktor Prasolov & Yuri Solovyev (1997) Elliptic Functions and Elliptic Integrals, page one, Translations of Mathematical Monographs volume 170, American Mathematical Society.
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http://crypto.stackexchange.com/questions/3807/dimension-of-encryption-of-a-linearly-dependent-set-of-plaintexts?answertab=votes
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# Dimension of Encryption of a linearly dependent set of plaintexts
Suppose I have an encryption scheme of the form $E(P)=C$ where $P$ is the plaintext and belongs in $\{0,1\}^N$, $C$ the ciphertext which belongs to $\{0,1\}^M$ and the encryption is performed under the key $K$ in $\{0,1\}^R$.
If I pick a subspace of messages $M_0$ and I perform encryption on this set, can I conclude anything about the dimension of $E(M_0)$ as a subspace of $\{0,1\}^M$.
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## 1 Answer
If $E$ was a random function from $\{0,1\}^N$ to $\{0,1\}^M$, then encrypting $n = |M_0|$ distinct messages with $E$ would be equivalent to picking $n$ independent random elements from $\{0,1\}^M$. If $n \ll M$, these random elements are with very high probability linearly independent, and thus span a subspace of dimension $n$; conversely, if $n \gg M$, with high probability they'll span the entire $M$-dimensional space.
In any case, of course, $\dim(E(M_0)) \le \min(n,M)$. If $n \approx M$, there's a non-negligible probability that the inequality is strict, but it decreases exponentially in either direction. (In particular, the probability that $\dim(E(M_0)) < n$ equals $1 - \prod_{i=0}^{n-1}( 1-2^{i-M} ) \le \sum_{i=0}^{n-1} 2^{i-M} < 2^{n-M}$.)
Now, determining the dimension of the subspace spanned by $E(M_0)$ is computationally fairly easy (at least unless both $M$ and $n$ are really huge), so any non-negligible difference in it between the actual encryption scheme $E$ and a random function would allow a distinguishing attack on $E$. Secure encryption schemes are not supposed to be distinguishable from random functions (without knowledge of the key), so if $E$ is secure, there shouldn't be any detectable statistical difference between the dimension of $E(M_0)$ and that predicted for a random function.
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http://physics.stackexchange.com/questions/47859/the-definition-of-an-inertial-reference-frame-in-einsteins-relativity/47872
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# The definition of an inertial reference frame in Einstein's relativity
I'm reading Sean Carroll's book on general relativity, and I have a question about the definition of an inertial reference frame. In the first chapter that's dedicated to special relativity, the author describes a way of constructing a reference frame in the following manner:
"The spatial coordinates (x, y, z) comprise a standard Cartesian system, constructed for example by welding together rigid rods that meet at right angels. The rods must be moving freely, unaccelerated. The time coordinate is defined by a set of clocks, which are not moving with respect to spatial coordinates. The clocks are synchronized in the following sense. Imagine that we send a beam of light from point 1 in space to point 2, in a straight line at a constant velocity c, and then immediately back to 1 (at velocity -c). Then the time on the coordinate clock when the light beam reaches point 2, which we label $t_2$, should be halfway between the time on the coordinate clock when the beam left point 1 ($t_1$) and the time on the same clock when it returned ($t^{'}_{1}$): $$t_2=\frac{1}{2}(t^{'}_{1}+t_1)$$ The coordinate system thus constructed is an inertial frame".
First of all, it is not completely clear what does "the rods must be moving freely, unaccelerated" exactly mean. Unaccelerated compared to what?
Secondly, and this is my main question, is the ability to synchronize clocks is unique to inertial frames? If the frame is not inertial, in the sense that Newton's second law $\vec{F}=\frac{d\vec{p}}{dt}$ does not hold, is it still possible that for a set of clocks which are not moving with respect to the spatial coordinates of this frame, that the equation $t_2=\frac{1}{2}(t^{'}_{1}+t_1)$ will always hold for any 2 points in space and a beam of light traveling between them? Can the ability to synchronize clocks be used as a criteria for inertial frames?
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Welcome Andrey. Good Question. I hope it is attend to shortly. – Chris Dec 29 '12 at 15:13
2
– twistor59 Dec 29 '12 at 17:34
1
Sean Carroll gives the appearence that the synchronization procedure is part of the construction. I think for an inertial frame it is sufficient to assert the rigidity of the rods and the non-acceleration. Then you want coordintes, for space and for time. The rods provide spatial coordinates, and you disseminate time to your clocks so that you're able to assign time coordinate consistently. I guess my point is: while you do need to disseminate time (synchronize the clocks), if the frame is non-accelerating it's an inertial frame anyway. – Cleonis Dec 29 '12 at 18:24
Take a look at my comment on accepted answer. – Sachin Shekhar Jan 8 at 5:28
## 4 Answers
What Sean Carroll refers to is acceleration as indicated by an accelerometer that is right next to the rods, co-moving with the rods.
The readout of an accelerometer is a local measurement. That is important in this stipulation about the rigid rods. The demand is not about being unaccelerated with respect to some other object that may be at some distance, it's about strap-on accelerometers giving a readout of zero.
Can the ability to synchronize clocks be used as a criteria for inertial frames?
For the synchronisation procedure to work (to not run into inconsistencies), the speed of light must be the same in all directions. As we know, that is the case only for an observer in inertial motion.
Addressing your question from a more general perspective:
Thought experiments involving clocks being synchronized are pretty much always scenarios where the clocks are a great distance apart. Light is so fast, you want a good bit of distance. On the other hand, in the context of GR, when you take an inertial frame of reference as conceptual starting point, the scenario is that you are thinking really locally.
The frame that is co-moving with the International Space Station as it orbits the Earth is such a local inertial frame of reference. When you zoom out the next level is the inertial frame that is co-moving with the Earth's center of mass. And so you zoom out to ever larger perspectives.
In the GR concept of inertial frame of reference the inertial frames of each of those levels of perspective are in motion relative to each other. That is why as a starting point you start with a concept of a local inertial frame of reference.
Your question is not wrong, but I can't think of any useful thought experiment that features the combination that you ask about: the synchronization procedure, and the distinction between inertial and non-inertial frame.
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This is wrong. Accelerometer would tell a freely falling reference frame that its accelerating when actually its not. A freely falling reference frame is an inertial reference frame in General Relativity. – Sachin Shekhar Jan 8 at 5:27
@Shekhar OK, let me state explicitly. With 'accelerometer' I mean the type of acceleration sensor that is in many of today's smartphones. When you throw such a smartphone up in the air then during its flight the smartphone's accelerometer will read zero. (Or very close to zero, if the accelerometer is sensitive enough to detect air drag.) So I mean detection of free fall motion. By contrast: a GPS based navigation device can evaluate successive positions and then compute acceleration with respect to some reference. Clearly I do not mean that type. Please correct your comment accordingly. – Cleonis Jan 9 at 22:14
Which type of experiment you performed with by throwing your smartphone up in the air... Simply drop it and read the values. All accelerometers are same despite there can be sensitivity differences. – Sachin Shekhar Jan 9 at 22:34
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The standard test to find out if you are in an inertial frame is to surround yourself with some non-interacting particles e.g. in a sphere. If the shape made up by the particles does not change with time then you are in a (locally at least) inertial frame. Curved space can be detected by either the volume and/or the shape made by the particles changing.
Bearing this in mind, in the example you give the rods are unaccelerated if they stay in the same position relative to each other i.e. the non-acceleration is with respect to each other and indeed everything else sharing their inertial frame.
Re your question about clocks, it isn't possible to synchronise clocks in different frames because they run at different rates. Well, I suppose it's possible to synchronise them once, but they won't stay in synch. In GR the clocks don't even need to be moving wrt each other. If I hover above a black hole and lower a clock on a rope to near the event horizon, that clock and my clock will run at different rates even though they aren't moving relative to each other.
In GR we don't really use the idea of inertial frames much. Instead we talk about spacetime being locally flat (Minkowski). In this sense it's possible to be in an inertial frame even when accelerating. The inside of the ISS is approximately an inertial frame even though it's accelerating towards the Earth at a fair clip.
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I didn't meant synchronizing clocks in different frames. Suppose I'm on spaceship accelerating at a constant acceleration relative to some inertial frame. A passenger could easily tell that the spaceships's reference frame is not inertial because Newton's second law doesn't hold, e.g for a pendulum like Vladimir proposed. But can we put a set of clocks at each point of the spaceship like Caroll described so that $t_2=\frac{1}{2}(t_1+t^{′}_{1})$ will always hold for light? – Andrey B Dec 29 '12 at 17:30
3
@ Andrey B - No, inconsistencies will build up, and they build up in more ways than one. As observed from a non-inertial frame the speed of light is not the same in all directions. Worse: in an accelerating spaceship the clocks in front are not in the same frame as the clocks in the rear. The effect is more pronounced with two spaceships moving co-linear, and both accelerating. Even when they both have the same proper acceleration their clocks cannot be synchronized consistently. Similar problem with a single spaceship for clocks in the front and the rear. – Cleonis Dec 29 '12 at 17:32
OK than, this answers my question. – Andrey B Dec 29 '12 at 17:58
First of all, it is not completely clear what does "the rods must be moving freely, unaccelerated" exactly mean. Unaccelerated compared to what?
Nice Question. I'd like to add: Moving freely with respect to what?
Well, its somewhat conventional. Newtonian mechanics defined an absolute inertial reference frame attached with heliocenter of Sun for that purpose. Einstein didn't defined such things, but borrowed one of conventionally defined inertial reference frames to define new ones unless new ones are available for defining purpose.
It means, the rods must be moving freely, unaccelerated with respect to another inertial reference frames. It may not look plausible, but an inertial reference frame can't be defined alone.
If the frame is not inertial, in the sense that Newton's second law $\vec{F}=\frac{d\vec{p}}{dt}$ does not hold, is it still possible that for a set of clocks which are not moving with respect to the spatial coordinates of this frame, any 2 points in space and a beam of light traveling between them, that the equation $t_2=\frac{1}{2}(t^{'}_{1}+t_1)$ will always hold?
We are defining things here, not validating/verifying things. Clocks can be synchronized with the same rule in non-inertial reference frame as long as light reaches those points (in some of non-inertial cases, light can fail to reach another point due to change of local geometry of Spacetime).
Is the ability to synchronize clocks all over space is unique to inertial frames? Can the ability to synchronize clocks be used as a criteria for inertial frames?
No. Many of non-inertial cases can synchronize clocks with the same rule. As there is no absolute/universal clock, you can't really comment on those synchronizations (simply relativistic physics isn't that powerful for that comment).
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Acceleration is an absolute thing. Unaccelerated means that, say, a pendulum attached to these rods does not move at all at any time. Acceleration of rods would make it deviate depending on acceleration, like a passenger in a car. Clock synchronization can then be made in a "simple" way described by Sean.
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http://crypto.stackexchange.com/questions/6303/elgamal-signature-forging-a-signature-of-a-specific-form
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# ElGamal signature: Forging a signature of a specific form
I have a question I can't solve from one of the courses I'm currently taking:
Show that given a legitimate ElGamal signature $(S,R)$ on a given message $m$, an attacker
can compute a signature on messages of the form: $m'=(m+bS)a\mod p - 1$,
when $b\in Z_p^*$ is chosen as the attacker wishes and $a=g^b\mod p$.
Hint: Observe the value of $g^{ma+baS}$
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http://www.physicsforums.com/showthread.php?t=211934
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Physics Forums
## some higher order differential equations
1. The problem statement, all variables and given/known data
a) s^2*t''+st*t'=s
b) y(dx/dy)^2=x^2+1
c) 17y''''-t^6*y"-4.2y^5=3cost(t)
2. Relevant equations
3. The attempt at a solution
For part a I thought about doing a reduction of order but I can't because I have the variable s present. Not sure what my other options are.
Part b I solved for dx/dy but now I'm stuck. Can't think clearly
Part c I'm not even remotely sure on how to do this. I was thinking integrating factor but I don't know.
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Recognitions: Gold Member Science Advisor Staff Emeritus You can solve (2) algebraically for dx/dy: $$\frac{dx}{dy}= \pm \frac{\sqrt{x^2+ 1}}{\sqrt{y}}$$ and have two separable differential equations. As for the other two, they are badly non-linear and I have no clue!
If the 1st one is $$s^2\,t''+s\,t\,t'=s\Rightarrow t\,x''+x\,x'=1$$ then as HallsofIvy said it is non-linear. (I changed the notation a bit). The general strategy for these ODE's are to search for Lie point symmetries. Unfortunately this one does not have any A second approach is to search for a first integral of the equation, which is based on a search and a try method. For this ODE if you try a first integral of the type $$I=f(x,t)\,x'+g(x,t)=C$$ you can find that $$I=t\,x'+\frac{1}{2}x^2-x-t=C$$ which is a Riccati type 1st order ODE. With the standard transformation you can convert it in to a 2nd order homogeneous linear ODE which can be solved the help of Bessel functions.
## some higher order differential equations
By carefully following Rainbow Child's instructions I think I was able to solve the first equation. The first part was to find the first integral, which was indeed as stated:
$$f(x,t)\cdot x' +g(x,t)=C$$
The functions f and g can be found by taking the derivative with respect to t giving:
$$fx''+f'x'+g'=0$$
Setting this equal to the original equation gives then:
$$f(x,t)=t$$
and thus f'=0, and also
$$g(x,t)=\frac{x^2}{2}-x-t$$
We have now:
$$t\cdot x'+ \frac{x^2}{2}-x-t=C$$
or:
$$\frac{dx}{dt}-\frac{1}{t}\cdot x + \frac{1}{2t}x^2=\frac{C}{t}+1$$
The Riccati equation:
$$\frac{dx}{dt}+Q(t)x+R(t)x^2=P(t)$$
can be transformed by using:
$$x=\frac{u'}{Ru}$$
Doing this transforms the equation into:
$$t^2\frac{d^2u}{dt^2}-\frac{1}{2}(t+C)u=0$$
Which is a special case of the following extended Bessel equation:
$$t^2\frac{d^2u}{dt^2}+(2k+1)t\frac{du}{dt}- \left(\alpha^2t^{2r}+\beta^2 \right)u=0$$
Which has the solution:
$$u(t)=t^{-k}\left[AI_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) + BK_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) \right]$$
with:
$$\gamma=\sqrt{k^2-\beta^2}$$
The functions are the modified Bessel functions of the first and second kind of order $$\frac{\gamma}{r}$$.
In our case we have:
$$k=-\frac{1}{2} \qquad r=\frac{1}{2} \qquad \beta^2=\frac{C}{2} \qquad \alpha^2=\frac{1}{2}$$
and:
$$\frac{\gamma}{r}=\sqrt{1-2C}=n \qquad \frac{\alpha}{r}=\sqrt{2}$$
The solution is thus:
$$u(t)=\sqrt{t} \left[ A I_{n}\left(\sqrt{2t}\right) + B K_{n}\left(\sqrt{2t}\right) \right]$$
After transforming back to x(t) we get:
$$x(t) = 1+\sqrt{2t} \left[ \frac{\displaystyle \frac{A}{2} \left[I_{n-1}\left(\sqrt{2t}\right) + I_{n+1}\left(\sqrt{2t}\right) \right]- \frac{B}{2} \left[K_{n-1}\left(\sqrt{2t}\right) +K_{n+1}\left(\sqrt{2t}\right) \right]}{\displaystyle A I_{n}\left(\sqrt{2t}\right) + B K_{n}\left(\sqrt{2t}\right)} \right]$$
Rewriting a little bit gives now:
$$x(t) = 1+\sqrt{\frac{t}{2}} \left[ \frac{\displaystyle \left[I_{n-1}\left(\sqrt{2t}\right) + I_{n+1}\left(\sqrt{2t}\right) \right]- D \left[K_{n-1}\left(\sqrt{2t}\right) +K_{n+1}\left(\sqrt{2t}\right) \right]}{\displaystyle I_{n}\left(\sqrt{2t}\right) + D K_{n}\left(\sqrt{2t}\right)} \right]$$
This solution has indeed two parameters C (in n) and D. To see if it is the correct one, it must be validated using the original equation. I haven't done this yet, the algebra is somewhat involved.
Non-linear differential equations are indeed not easy. Thanks to Rainbow Child for the instructions, it was a pleasure doing this calculation.
@ roldy: Is this the result you were looking for? Where exactly do these equations come from?
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http://www.physicsforums.com/showthread.php?p=4056508
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Physics Forums
Intersection of concave functions
Hi all,
I have a question. Suppose f : [ 0, l) $\rightarrow$ ℝ is concave , increasing and continuous where l < ∞ and g : [ 0, l) $\rightarrow$ ℝ is also concave, nondecreasing and continous on the same interval. Can we claim that f and g intersect finitely many times in this interval (possibly 0) ? What if number l replaces with infinity?
Thanx in advance, H.
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Recognitions: Science Advisor This looks a bit like a homework question, so I'm not going to just give you the answer! The important thing here is that the functions are defined on [0,1), not on [0,1]. In other words they can tend to infinity as ## x \rightarrow 1##. EDIT: Sorry, I misread "concave" as "convex" here.
Recognitions: Science Advisor Maybe it is helpful to say that if f=g infinitely-often, then h=f-g has infinitely-many zeros in [0,l). Only continuous ,monotone function with infinitely-many zeros I can think off is a function of the type d(x,S) , i.e., the distance function between a point and a set. Edit: there is a result that every closed set is the zero set of a smooth function -- more so a continuous one, so there are a lot of options for h=f-g. Also, if the functions are monotone, then they are a.e. differentiable, so it may make sense to assume differentiability to see what happens. Let me think it through some more, tho.
Intersection of concave functions
Quote by AlephZero This looks a bit like a homework question, so I'm not going to just give you the answer! The important thing here is that the functions are defined on [0,1), not on [0,1]. In other words they can tend to infinity as ## x \rightarrow 1##.
Alephero, can you explain that last line (maybe give reference )? Also I think here being increasing is also important , I think only concaveness is not enough here.
and it's not a homework question but I need it as a part of an bigger argument :)) . AlephZero, your last line made me wonder : As far as I get, the domain is also important here. What if f, g : [0, ∞ ) $\rightarrow$ ℝ, can we claim they intersect finitely many times in [0,1) ? Here I guess there's no problem with right end point 1.
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Quote by hermanni Alephero, can you explain that last line (maybe give reference )? Also I think here being increasing is also important , I think only concaveness is not enough here.
For example take f(x) = 1/(1-x).
Define g(x) by an infinite set of straight line segments joined end to end on the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ...
You can make f(x) and g(x) intersect twice within each interval.
That type of example doesn't work if f(x) is defined on the closed interval [0,1] , because you can't define "##f(1) = \infty##".
You can make the same idea work for the interval [0, ∞ ). For example take. with ##f(x) = x^2##. If f(0) and f(1) ar both finite, this idea for getting an infinite number of intersections doesn't work. The theorem that a continuous function on a closed interval is uniformly continuous will probably come into a proof that there are only a finite number of intersections if f(0) and f(1) are both finite.
EDIT: Again, sorry, this is about convex functions not concave.
Quote by AlephZero For example take f(x) = 1/(1-x). Define g(x) by an infinite set of straight line segments joined end to end on the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ... You can make f(x) and g(x) intersect twice within each interval. That type of example doesn't work if f(x) is defined on the closed interval [0,1] , because you can't define "##f(1) = \infty##". You can make the same idea work for the interval [0, ∞ ). For example take. with ##f(x) = x^2##. If f(0) and f(1) ar both finite, this idea for getting an infinite number of intersections doesn't work. The theorem that a continuous function on a closed interval is uniformly continuous will probably come into a proof that there are only a finite number of intersections if f(0) and f(1) are both finite.
OK, my functions f and g are defined on [0, ∞) and I needed finiteness of number of intersections in [0, 1). They're both bounded, concave, increasing and BOUNDED on [0, ∞). I have now boundedness, can I claim finite # of intersections on [0,1)? I think I can do it for [0,1] based on what you said, here does endpoint 1 matter now?
Recognitions: Homework Help Science Advisor Are we all using the same definition of concave? The definition I'm used to is the one at http://en.wikipedia.org/wiki/Concave_function. According to that, 1/(1-x) (on [0,1)) would be convex. hermanni, please clarify.
Quote by haruspex Are we all using the same definition of concave? The definition I'm used to is the one at http://en.wikipedia.org/wiki/Concave_function. According to that, 1/(1-x) (on [0,1)) would be convex. hermanni, please clarify.
Well, I'm using the same definition and you're right, I didn't read carefully and 1/1-x is convex.
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Quote by haruspex Are we all using the same definition of concave? The definition I'm used to is the one at http://en.wikipedia.org/wiki/Concave_function. According to that, 1/(1-x) (on [0,1)) would be convex.
Oops. So ignore my comments about the open and closed intervals!
Bu I don't think this changes my assertioon that there can be an infinite number of intersections, with the conditions in the OP. Given a smooth enough function f(x) you can still construct a function g(x) from an infiite number of line segments, with an infinite number of intersections.
Of crosue the OP's functions may have some more properties we don't know about, which prevent this.
Quote by AlephZero Oops. So ignore my comments about the open and closed intervals! Bu I don't think this changes my assertioon that there can be an infinite number of intersections, with the conditions in the OP. Given a smooth enough function f(x) you can still construct a function g(x) from an infiite number of line segments, with an infinite number of intersections. (*)Of crosue the OP's functions may have some more properties we don't know about, which prevent this.
OK, I got the idea of construction, thank you very much.
(*) Do you think boundedness is one of these properties? If so, how can I show?
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Quote by hermanni Do you think boundedness is one of these properties?
I don't think boundedness gets you far. Here's another construction:
Consider the points (1-1/n, 1-1/n2), n = 1, 3, 5... Join the dots with straight lines. Now do the same with n = 2, 4, 6... And smoothness is not going to help - it wouldn't be hard to smooth out the corners.
Conditions on higher derivatives might do it, but I'd be surprised.
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Quote by AlephZero For example take f(x) = 1/(1-x). Define g(x) by an infinite set of straight line segments joined end to end on the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ... You can make f(x) and g(x) intersect twice within each interval. That type of example doesn't work if f(x) is defined on the closed interval [0,1] , because you can't define "##f(1) = \infty##". You can make the same idea work for the interval [0, ∞ ). For example take. with ##f(x) = x^2##. If f(0) and f(1) ar both finite, this idea for getting an infinite number of intersections doesn't work. The theorem that a continuous function on a closed interval is uniformly continuous will probably come into a proof that there are only a finite number of intersections if f(0) and f(1) are both finite. EDIT: Again, sorry, this is about convex functions not concave.
I see, so the idea is that the function/curve is rectifiable, right?
How is f(x)=1/(1-x) non-decreasing?
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http://mathhelpforum.com/calculus/96289-notes-wrong.html
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# Thread:
1. ## Are the notes wrong?
I've highlighted parts of the image I'm concerned with:
In other words, they want a cubic polynomial, i.e. $P_3$, and if they say that a given polynomial's degree is $n+1$, they must choose $n=2$ so that $n+1=3$.
Perhaps they should have written the degree of the poly is $n$, not $n+1$?
Something here seems inconsistent to me.
2. There doesn't seem to be any inconsistency. You need to use the zeros of the (n+1)th degree Chebshev polynomial in order to get an n'th degree polynomial approximation for a given function f(x).
Notice that a polynomial of degree n has n+1 coefficients (because of the constant term). You need to sample f(x) at n+1 points in order to get sufficient data to determine these coefficients. At the very simplest level, you need to sample the function at 2 points in order to get a straight line (i.e. degree 1) approximation for it.
3. Thanks. I sort of figured this a while after posting, but your post confirmed my thoughts.
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http://math.stackexchange.com/questions/74875/dual-space-of-the-space-of-finite-measures
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# Dual space of the space of finite measures
Since I am reading some stuff about weak convergence of probability measures, I started to wonder what is the dual space of the space consisting of all the finite (signed) measures (which is well known to be a Banach space with the norm being total variation). Is there any characterization of it? We may impose extra assumptions on the underlying space if necessary.
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2
– t.b. Oct 22 '11 at 17:41
@t.b.: Thanks for sharing. Perhaps the one GEdgar gives below is one the "various sorts of representations" they mentioned in the footnote? – Syang Chen Oct 23 '11 at 3:34
yes, I think so. It is unsatisfactory in that you can't say what you really get. Something humungous, in any case. – t.b. Oct 23 '11 at 3:42
## 1 Answer
Well, your space of measures is isometric to $L^1(\mu)$ for some (probably very big, non-sigma-finite) measure $\mu$. So it is enough to know what is the dual of an $L^1$ space.
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1
Thanks, could you sketch how the "big space" is obtained? Or could you give some reference? – Syang Chen Oct 22 '11 at 20:42
Take a maximal family $\mathcal A$ of mutually singular probability measures. (Use Zorn's Lemma.) The space of measures is isometrically the $l_1$-sum of $L^1(\nu)$ as $\nu$ ranges over the family $\mathcal A$. – GEdgar Oct 22 '11 at 22:13
So there is no canonical representation for it? What's dual of $L^1$ when the space is not $\sigma$-finite? – Syang Chen Oct 23 '11 at 4:03
@Xianghong: Note that you have an $l_1$-sum of $L^1(\nu)$ where $\nu$ are probabilities. Its dual space is the $l_\infty$-product of the corresponding $L^\infty(\nu)$-spaces. – t.b. Oct 23 '11 at 10:02
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http://crypto.stackexchange.com/questions/2720/brute-force-a-ciphered-message/2724
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# Brute force a ciphered message?
I wrote my own cipher to encrypt messages. I would like to test a sample ciphered message to see how strong it is. Are there any tools for such task either in Windows or Linux ?
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## 2 Answers
There's no way to verify the strength of a cipher in any automated way, especially given only the ciphertext. Attacking the construction of a cipher is an inherantly intellectual process, which does not fit well to the qualities of computer analysis.
For example, I could write a cipher such that $f(m) = rand(0,n)$ where $n$ is the largest integer representable in $sizeof(m)*8$ bits, and any automated analysis would say "this is an awesome cipher", but a human would immediately realise that it's totally useless.
If you're writing your own ciphers for real usage, I have only one piece of advice: don't.
Creating your own ciphers is fun, and it can give you a learning experience if you're actually doing it based on constructions you understand, but the resulting cipher should never ever go anywhere near a production system. Experts put their ciphers through years (decades, even) of peer review before they're adopted by anyone. If your goal is to learn, do so by breaking ciphers.
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To calculate the time to brute force, you actually just have to look at the (effective) key space. If there are $k$ different possible keys, and all of them encrypt differently¹, it takes in average about $k/2$ tries to brute-force your cipher. How long this takes then depends on the speed of your cipher (or an optimized version thereof), and the hardware resources of an attacker.
¹This seems to be trivially the case, but it isn't: While DES uses formally a 64-bit key, 8 of these bits are not used, thus the effective key size is 56 bits, which means that in average $2^{56}/2 = 2^{55}$ tries are enough for brute-force.
Of course, with home-brewn ciphers usually the best attack is not a brute-force attack, but some clever cryptanalysis, which then only needs way less time. These ciphers are then called "broken".
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http://mathoverflow.net/questions/68041/showing-block-diagonal-structure-of-matrix-by-reordering
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## Showing block diagonal structure of matrix by reordering
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose we have a block-diagonal matrix $M$, but the block diagonal structure is not immediately apparent from looking at the matrix because the rows/columns are shuffled.
I wish to find a reordering of rows and columns, $M' = P M P^{-1}$, where $P$ is a permutation matrix, that will make the block structure apparent.
When this problem is solved, I wish to do the same for the case when the matrix does not have a perfect block structure, but there are a few non-zero elements scattered here and there.
One idea is to look at the eigenvectors, and the position of non-zero elements in each eigenvector will show the elements (i.e. row/column indices) belonging to t he respective blocks. Is this correct?
Any pointers are welcome. I'm sure there must be an easy solution to this.
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But... if a matrix has some nonzero elements scattered here and there, its structure is undistinguishable from the structure of any other non-block matrix... I think the question should be clarified a bit. – Qfwfq Jun 17 2011 at 12:14
Just imagine that you write down a block-diagonal matrix and add some extra elements. Of course by looking at this thing one can tell that it is an "almost" block-diagonal matrix. I am looking for something that captures this intuition. Coming up with a precise definition is part of the problem. – Szabolcs Jun 17 2011 at 12:48
## 5 Answers
If the matrix corresponds to a $d$-regular graph, then one can make some easy observations. For instance, if it is a block matrix with $k$ components, then there will be a $k$-dimensional eigenspace of vectors with eigenvalue $d$. As you say, this gives significant information about the blocks. You can't say that an arbitrary eigenvector will pick out a component, but you can say that the eigenspace will be generated by characteristic functions of components, and the eigenvectors will be constant on the components. One would have to make the notion of approximation precise, but presumably there are reasonable notions of approximation such that if a matrix is approximately a block matrix with $k$ components and is still $d$-regular, then there are several vectors $v$ that approximately satisfy $Av=dv$ and that are approximately constant on the approximate components. And finally, if all you're interested in is the components, then probably it's not too important what the values of the matrix are (as long as they are clearly zero or clearly non-zero), so perhaps you can reweight them so as to reduce to something like the regular case -- but I'm less sure about that last idea. All this would of course rely on points within components being significantly more connected than points in different components.
Another, related, way of detecting this would be to look at powers of the matrix, which would fill up the approximate components much more quickly than the links between components. In fact, I'd be tempted to pick a point and apply the matrix a few times to that point (considered as a unit vector) in order to identify its "approximate component". In the exact case, the component would consist of all points where the value is non-zero, whereas in the approximate case one would have to go for some kind of cutoff. Then one could repeat for other points. The resulting "components" might overlap a bit, but one could then just arbitrarily assign points in the overlap to some component or other.
I'm not sure what to make of your suggestion that coming up with a precise definition is part of the problem: I think there are probably several reasonable precise definitions, each of which leads to a different version of the problem. So either you should be more specific about why you want to do this or you should just choose your method of proof and define the problem in such a way that the method works. (So, for instance, you might decide on the above cutoff method and then choose a notion of "approximate block matrix" that guarantees that that process approximately identifies -- in some other sense that you make precise -- the approximate blocks.)
This kind of question is of interest in additive combinatorics. If $E$ is a set of integers, one can define a matrix $A$, indexed by $E\times E$, where $A(x,y)$ is the number of pairs $(z,w)\in E^2$ such that $x-y=z-w$. It would be nice to have a good way of splitting $E$ up into "approximate components" of this matrix/weighted graph, which would be "approximate additive components" of $A$. Even nicer would be to be able to say something nice about the additive structure of these components, on the assumption that there are at least $c|E|^3$ quadruples $(x,y,z,w)\in E^4$ with $x-y=z-w$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If you make believe $M$ is the adjacency matrix of a graph (vertices $i$ and $j$ are adjacent if and only if at least one of the two entries $m_{ij},m_{ji}$ is nonzero), then there are efficient algorithms for finding the connected components of the graph, which should correspond to the blocks of the matrix.
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Indeed. But the interesting case is when is not perfectly diagonal, i.e. there are a few links between those components, but not too many. – Szabolcs Jun 17 2011 at 12:45
Given a symmetric matrix $M\in\mathbb{R}^{n\times n}$ you are able to visualize the matrix as an adjacency matrix of a graph. Where node $i$ and $j$ have an edge between them if the $(i,j)$ entry of $M$ is non-zero. When the matrix is not symmetric then you end up looking at a directed graph.
A $k$ clustering algorithm on the associated graph should do a good job. Where $k$ is the number of blocks you want. This should also work when the matrix is not completely block diagonal.
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You might want to have a look at the Cuthill-KcKee algorithm and from there perhaps also chase the other sparse matrix reordering algorithms. The abovecited method seeks to minimize the bandwidth of a sparse matrix---so, although it does not produce a block-diagonal, it produces a banded matrix (with narrow bandwidth), which might be useful to you.
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First we set $\tilde M$ for the matrix obtained by replacing all non-zero elements by 1. Start with a vector $\tilde v_1$ having exactly one coordinate equal to $1$ and all other coordinates zero. Consider $v_2=v_1+\tilde Mv_1$ and define $\tilde v_2$ by replacing all non-zero elements (which are in fact strictly positive integers) of $v_2$ with $1$. Iterate until you get $\tilde v_{k+1}=\tilde v_k$. The coordinates of $\tilde v_k$ form then a block. Removal of this block and iteration of this algorithm gives the block decomposition. (This is more or less one of the methods suggested by Gowers and I guess that it is also related to the Cuthill-McKee algorithm mentioned by Suvrit.)
Correction: The above algorithm does not give decomposition into blocks but a decomposition into blocks on the diagonal plus upper triangular stuff.
One has thus also to check if later appearing blocks interact with blocks considered previously. This can be done by considering an upper triangular matrix with rows and columns indexed by the constructed diagonal blocks. Its coefficients are $1$ on the diagonal and $1$ for indices $i,j$ such that the image of of the $j-$th block involves the subspace spanned by the $i-$th block (where we consider the decomposition $V=V_1\oplus V_2\oplus\dots$ with $V_i$ corresponding to the image spanned by the $i-$th diagonal block).
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In essence, this also corresponds to interpreting the matrix as a graph, then starting from a node, visiting all nodes that can be visited, thus finding the largest connected component containing the starting node. Is this correct? – Szabolcs Jun 18 2011 at 9:25
Yes, except that I forgot to take into account nilpotency properties, now fixed. – Roland Bacher Jun 18 2011 at 10:21
Oops, good you pointed that out – Szabolcs Jun 18 2011 at 10:41
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http://mathoverflow.net/revisions/44789/list
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## Return to Answer
3 Fixed errors in the references...
If $A$ is symmetric, then the matrices that you mention are called:
Conditionally positive definite (CPD) --- these are intimately related to the venerable infinitely divisible matrices
There is a vast amount of literature on these matrices, some useful pointers can already be found in R. Bhatia's wonderful book: Positive definite matrices
There are some basic algorithmic approaches to check whether a matrix is CPD or not (e.g., Ref. 3 below)
A simple characterization is given by the following. Let $A$ be an $n \times n$ Hermitian matrix, and let $B$ be the $(n-1) \times (n-1)$ matrix with entries
$$b_{ij} = a_{ij} + a_{i+1,j+1} - a_{i,j+1} - a_{i+1,j}$$
Then $A$ is CPD if and only if $B$ is positive-definite.
References
1. R. Bhatia. Positive definite matrices (esp. Chapter 5)
2. R. B. Bapat and T. E. S. Raghavan. Nonnegative matrices and applications (Chapter 4)
3. Kh. D. Ikramov and N. V. SaveFevaSavel'eva. Conditionally positive definite matrices, J. Mathematical Sciences, Vo. 98, No. 1, 2000.
4. R. A. Horn. The theory of infinitely divisible matrices and kernels (e.g. here : http://www.ams.org/journals/tran/1969-136-00/S0002-9947-1969-0264736-5)/S0002-9947-1969-0264736-5.pdfhttp://www.ams.org/journals/tran/1969-136-00/S0002-9947-1969-0264736-5/S0002-9947-1969-0264736-5.pdf)
2 added symmetry as a prereq
The
If $A$ is symmetric, then the matrices that you mention are called:
Conditionally positive definite (CPD) --- these are intimately related to the venerable infinitely divisible matrices
There is a vast amount of literature on these matrices, some useful pointers can already be found in R. Bhatia's wonderful book: Positive definite matrices
There are some basic algorithmic approaches to check whether a matrix is CPD or not . Once I find the paper in my archives, I will update my answer(e.g., Ref. 3 below)
A simple characterization is given by the following. Let $A$ be an $n \times n$ Hermitian matrix, and let $B$ be the $(n-1) \times (n-1)$ matrix with entries
$$b_{ij} = a_{ij} + a_{i+1,j+1} - a_{i,j+1} - a_{i+1,j}$$
Then $A$ is CPD if and only if $B$ is positive-definite.
References
1. R. Bhatia. Positive definite matrices (esp. Chapter 5)
2. R. B. Bapat and T. E. S. Raghavan. Nonnegative matrices and applications (Chapter 4)
3. Kh. D. Ikramov and N. V. SaveFeva. Conditionally positive definite matrices, J. Mathematical Sciences, Vo. 98, No. 1, 2000.
4. R. A. Horn. The theory of infinitely divisible matrices and kernels (e.g. here : http://www.ams.org/journals/tran/1969-136-00/S0002-9947-1969-0264736-5)/S0002-9947-1969-0264736-5.pdf
1
The matrices that you mention are called:
Conditionally positive definite (CPD) --- these are intimately related to the venerable infinitely divisible matrices
There is a vast amount of literature on these matrices, some useful pointers can already be found in R. Bhatia's wonderful book: Positive definite matrices
There are some basic algorithmic approaches to check whether a matrix is CPD or not. Once I find the paper in my archives, I will update my answer.
A simple characterization is given by the following. Let $A$ be an $n \times n$ Hermitian matrix, and let $B$ be the $(n-1) \times (n-1)$ matrix with entries
$$b_{ij} = a_{ij} + a_{i+1,j+1} - a_{i,j+1} - a_{i+1,j}$$
Then $A$ is CPD if and only if $B$ is positive-definite.
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http://mathoverflow.net/revisions/13903/list
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## Return to Answer
1 [made Community Wiki]
Tarski ran into some trouble when he tried to publish his result that the Axiom of Choice is equivalent to the statement that an infinite set $X$ has the same cardinality as $X \times X$.
From Mycielski:
"He tried to publish his theorem in the Comptes Rendus but Frechet and Lebesgue refused to present it. Frechet wrote that an implication between two well known propositions is not a new result. Lebesgue wrote that an implication between two false propositions is of no interest. And Tarski said that after this misadventure he never tried to publish in the Comptes Rendus."
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http://mathoverflow.net/questions/64365/natural-transformations-as-categorical-homotopies/75686
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## Natural transformations as categorical homotopies
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Every text book I've ever read about Category Theory gives the definition of natural transformation as a collection of morphisms which make the well known diagrams commute. There is another possible definition of natural transformation, which appears to be a categorification of homotopy:
given two functors $\mathcal F,\mathcal G \colon \mathcal C \to \mathcal D$ a natural transformation is a functor $\varphi \colon \mathcal C \times 2 \to \mathcal D$, where $2$ is the arrow category $0 \to 1$, such that $\varphi(-,0)=\mathcal F$ and $\varphi(-,1)=\mathcal G$.
My question is:
why doesn't anybody use this definition of natural transformation which seems to be more "natural" (at least for me)?
(Edit:) It seems that many people use this definition of natural transformation. This arises the following question:
Is there any introductory textbook (or lecture) on category theory that introduces natural transformation in this "homotopical" way rather then the classical one?
(Edit2:) Some days ago I've read a post in nlab about $k$-transfor. In particular I have been interested by the discussion in the said post, because it seems to prove that the homotopical definition of natural transformation should be the right one (or at least a slight modification of it). On the other end this definition have always seemed to be the most natural one, because historically category theory develop in the context of algebraic topology, so now I've a new question:
Does anyone know the logical process that took Mac Lane and Eilenberg to give their (classical) definition of natural transformation?
Here I'm interested in the topological/algebraic motivation that move those great mathematicians to such definition rather the other one.
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"anybody" does use this definition (you can find it somewhere on my [pretty much defunct] blog, for example). – Todd Trimble May 9 2011 at 10:38
Wow, this is a great way to think about natural transformations! I wish I'd seen this months ago – David White Jul 1 2011 at 15:02
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I am happy to give you my 300th vote for this question. It is not the way natural transformation are introduced, but it is actually the way people (should) think about them. IMHO this is the starting observation to make for introducing simplicial categories as a model for $\infty$-categories. – DamienC Sep 8 2011 at 15:14
@DamienC I really appreciate your comment, I don't know very much about model categories so I'm wondering: "would you like to elaborate your comment in an answer in which discuss more completely this aspect of this definition of natural transformation?" (I think this argument is interesting and deserves to be in an answer) – Giorgio Mossa Sep 8 2011 at 16:08
I'll try to find some time to do this next week. But I can't promise (I'll soon have teaching duties, and I still have to work on preparing my lectures). – DamienC Sep 9 2011 at 14:03
show 3 more comments
## 8 Answers
The homotopy analogue definition of natural transformations has been known and used regularly since at least the late 1960's, by which time it was understood that the classifying space functor from (small) categories to spaces converts natural transformations to homotopies because it takes the category $I=2$ to the unit interval and preserves products. Composition of natural transformations $H\colon A\times I\to B$ and $J\colon B\times I\to C$ is just the obvious composite starting with $id\times \Delta: A\times I \to A\times I\times I$, just as in topology. (I've been teaching that for at least several decades, and I'm sure I'm not the only one.)
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Once you learn a subject, you can think about things in whatever way is most pleasing or helpful for solving a problem. Fixing a fact as a definition is pedagogy -- something to help those learning the subject.
I can't really speak for how others learn, but I'm not sure recognizing natural transformations as being described by functors $\mathcal{C} \times 2 \to \mathcal{D}$ would be very useful before one starts seriously thinking in terms of the 2-category of categories.
I confess I would almost turn your question on its head -- I far more frequently want to think of a homotopy between functions $f,g:X \to Y$ as being a function from $X$ to paths in $Y$, or sometimes as a function from $[0,1]$ to $Y^X$, and feel the usual definition as a function $X \times [0,1] \to Y$ more as being a much simpler way to state the technical details. I saw the analogy with homotopy early in learning about categories, and I don't think seeing natural transformations defined as functors $\mathcal{C} \times 2 \to \mathcal{D}$ would have helped me make the analogy. (But, for the record, I am very much not an algebraic topologist)
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I agree with you when you say that it's more natural thinking homotopy as a path of function, and in this sense one could think about natural transformation as functor of kind $2 \to \text{Fun}(\mathcal C,\mathcal D)$, by the way this require the category of functors and so natural transformation as well. A similar problem seems to arise in topology, where one must define a topology on the space of function in order to define homotopies like path of function. Why do you think this definition of natural transformation is useful when one start to think in terms of the 2-categories of categories? – Giorgio Mossa May 10 2011 at 20:02
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I assert that when one is thinking in terms of the 2-categories of categories, definitions are no longer important. (Amongst) what is important is the knowledge of various equivalent ways to capture a notion. I think describing natural transformations as functors $\mathcal{C} \times 2 \to \mathcal{D}$ doesn't become useful until thinking in terms of the 2-category of categories mainly because the only other way I see to use description is to unfold it into the ordinary description and working in terms of that. – Hurkyl May 11 2011 at 0:23
This "geometric" definition is well-known to category-theorists. See for example this youtube video by the Catsters, which introduces natural transformations. It should be also well-known to algebraic topologists working with model categories. But I have to admit that there are few introductions to category theory which emphasize this definition of a natural transformation.
Remark that this fits into a more general framework: For every category $C$, there is an isomorphism $[I,C] \cong Arr(C)$, where $Arr(C)$ is the arrow category of $C$. In particular, $Arr([C,D]) \cong [I,[C,D]] \cong [C \times I,D]$.
On the other hand, the usual definition is more easy to work with. For example how do you define the composition of two natural transformations, say given by $\alpha : C \times 2 \to D, \beta : C \times 2 \to D$ with $\alpha(-,1) = \beta(-,0)$? Of course you can just write it down explicitly, but then you end up working with the usual definition. But instead, you could also use that $\alpha,\beta$ correspond to a functor on the amalgam $(C \times 2) \cup_C (C \times 2)$ of the inclusions $(-,1)$ and $(-,0)$, and compose with the natural functor $C \times 2 \to (C \times 2) \cup_C (C \times 2)$ which "leaves out the middle point".
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I like this definition of composition of natural transformation, but it seems to use the notion of limit, which need the definition of natural transformation, am I right? I wrote down a definition of composition for this kind of natural transformation which is more complex of the classical one, but it has the merit of avoid to demonstrate that the composite is a natural transformation, because this fact is implicit in the definition. – Giorgio Mossa May 10 2011 at 20:17
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I guess the fastest answer is that the composition corresponds to (pulling back along) the canonical map $\vec I \to \vec I \cup_{\text{middle}} \vec I$, and leave the $C$s to the reader? – Theo Johnson-Freyd Sep 14 2011 at 17:28
Concerning
Have anyone ever introduced natural transformation in this "homotopical" way rather then the classical one in any reference like a textbook or some lecture notes?
Yes, Quillen introduces it in the paper
Higher Algebraic K-theory. I. Algebraic K-theory, I: Higher K-theories (Proc. Conf., Battelle Memorial Inst., Seattle, Wash., 1972), pp. 85–147. Lecture Notes in Math., Vol. 341, Springer, Berlin 1973.
In connection with his "Theorem A" and "Theorem B."
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Disclaimer: this is not an answer to the question as I have no explanation for why people don't introduce natural transformations in the way explained in the question, but I am posting this in order to expand a comment I made. The comment was
this is the starting observation to make for introducing simplicial categories as a model for $\infty$--categories
Moreover, I am not a specialist neither of category theory nor of homotopy theory (and a posteriori of higher categories).
## The $2$-category of categories
The starting point is that the category $Cat$ of categories is actually a $2$-category. For any to objects (i.e. categories) $\mathcal C$ and $\mathcal D$ we have that $Hom_{Cat}(\mathcal C,\mathcal D)$ is itself a category.
This is very transparent when using the definition $$Hom_{Cat}(\mathcal C,\mathcal D):=Hom_{t_{\leq0}(Cat)}(\mathcal C\times\Delta^1,\mathcal D)\,,$$ where $\Delta^1=\Box^1=\mathbb{G}^1$ is the arrow category $0\to 1$ and $t_{\leq0}(Cat)$ is the underlying $1$-category of $Cat$.
Remark: In general one can see a $2$-category $\mathcal C$ as a simplicial category by replacing the $Hom$-categories by their nerves.
In the case of $Cat$, we see that the $Hom$-categories naturally appear as $1$-truncations of simplicial sets (one can replace here "simplicial" by "cubical" of "globular").
## The $3$-category of $2$-categories
Le us now go to natural transformations of (strict) $2$-functors between (strict) $2$-categories. Given two such $2$-functors $F,G:\mathcal C\to\mathcal D$ one can see that a natural transformation $F\Rightarrow G$ is the same as a $2$-functors $$\phi:\mathcal C\times \mathbb{G}^2\to\mathcal D$$ such that $\phi(-,0)=F$ and $\phi(-,1)=G$, where $\mathbb{G}^2$ is the $2$-category with two objects $0$ and $1$ and such that $Hom_{\mathbb{G}^2}(0,1)$ is the arrow category $\mathbb{G}^1=(0\to 1)$.
Therefore the "set" of $2$-functors is a naturally a $2$-category.
Remark: as before we can then see any $3$-category as a simplicial/cubical/globular category by replacing the $Hom$-$2$-categories by their (simplicial/cubical/globular) nerves.
In the case of $2-Cat$, we see that the $Hom$-$2$-categories naturally appear as $2$-truncations of globular sets.
## Simplices, Cubes, and globes
The globe category $\mathbb{G}$, the cubical category $\Box$ and the simplicial category $\Delta$ are known to be suitable geometric shape to model higher structures. Simplicial sets are good models for (weak) $\infty$-groupoids. It was proved (by Jardine ... with some improvement by Cisinski if I remember well) that cubical sets also provide a model for (weak) $\infty$-groupoids.
I don't know any reference but I guess that the same holds for globular sets (which are quite more used by people working with automata).
## The $(n+1)$-category of $n$-categories
Let me consider the category $n-Cat$ of (strict) $n$-categories. A a natural transformation between (strict) $n$-functor $F,G:\mathcal C\to\mathcal D$ can be written as an $n$-functor $$\phi:\mathcal C\times \mathbb{G}^n\to\mathcal D$$ such that $\phi(-,0)=F$ and $\phi(-,1)=G$, where $\mathbb{G}^n$ is the $n$-category with two objects $0$ and $1$ and such that $Hom_{\mathbb{G}^n}(0,1)$ is the $(n-1)$-category $\mathbb{G}^{n-1}$.
Therefore the "set" of $n$-functors is a naturally a (strict) $n$-category, and thus $n-Cat$ is a (strict) $n+1$-category. It also naturally appears as a $n$-truncation of a globular category.
## The advantage of working with simplicial/cubical/globular categories
Working directly with simplicial/cubical/globular categories has the following advantages:
1. it does allow to work directly with higher categories without going through an inductive process.
2. it allows to deal with weak $(\infty,1)$-categories, as simplicial/cubical/globular are models for weak $\infty$-groupoids (here $(\infty,1)$ stands for "$\infty$-categories such that $n$-arrows for $n\geq2$ are weakly invertible").
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@DamienC thanks a lot, this stuff is really cool, though there are some things that I don't get straight, for instance: above you say that $\hom_{t \leq 0 (Cat)}(\mathcal C,\mathcal D)$ should be a category but it seems to me it should be a set, at the same time you say that a functor $\mathcal C \times G^n \to \mathcal D$ should be a $n$-functor, but in this case if we consider $n=2$, then a $2$-functor should be a natural transformation in the sense of the definition in the question, where am I wrong? – Giorgio Mossa Sep 11 2011 at 10:06
@Giorgio Mossa: $Hom_{t\leq0}(Cat)}(\mathcal C,\mathcal D)$ is a set but $Hom_{t\leq0}(Cat)}(\mathcal C\times(0\to 1),\mathcal D)$ is a category. About your second question you are right, I should have written that an $n$-functor $\mathcal C\times G^n\to\mathcal D$ is a natural transformation between two $n$-functors. I'm going to fix it. – DamienC Sep 11 2011 at 11:30
@DamienC But $Hom_{t≤0(Cat)}(\mathcal C\times \Delta^1,\mathcal D)$ should be the set of the functors between the categories $\mathcal C \times \Delta^1$ and $\mathcal D$, this should be (in our description) the set of natural transformations, this description seems lacking of the objects of the category, unless you're considering arrow-only categories. – Giorgio Mossa Sep 11 2011 at 11:59
@Giorgio Mossa: to be precise I should have said that `$Hom_{t\leq0}(Cat)}(\mathcal C,\mathcal D)$` is the set of morphisms of a category with objects being functors $\mathcal C\to\mathcal D$. – DamienC Sep 11 2011 at 19:55
@DamienC I suppose you meant $Hom_{t\leq0(Cat)}(\mathcal C \times \Delta,\mathcal D)$ in your last comment, but I think I get it. Thanks a lot. Just to be completely correct I think that there are some typo-error in your answer, you say that the natural-transformation $\phi$ should be such that $\phi(-,0)=F$ e $\phi(-,0)=G$, did you meant $\phi(-,1)=G$, right? – Giorgio Mossa Sep 11 2011 at 21:04
show 2 more comments
What is "more natural" is strictly determined by a mathematical background one has (or more seriously --- by one’s understanding of the world) when one comes to learn a new subject. Thus, a good definition should be more about "simplicity" (with respect to its theory) than about "analogy" to other concepts (in other braches of math). Analogies are then established by theorems.
I am not a mathematician, so I have a sweet opportunity to be ignorant on some fundamental branches of math --- for example --- topology. I think of functors $\mathbb{C} \rightarrow \mathbb{D}$ as of structures in $\mathbb{D}$ of the shape of $\mathbb{C}$. Then a transformation is something that morphs one structure into another (i.e. it is a collection of morphisms indexed by the shape of a structure), whereas natural transformation is something that morphs in a coherent way.
I really like a story on "Blind men and an elephant" link text that I first red in Peter Johnstone's book "Sketches of an Elephant". He compares a topos to the elephant, and we are the blind men. Surely, we are blind men, but I do think that most concepts found in category theory (with perhaps category theory itself) are like elephants.
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IMHO you are more "mathematician" than lots of my colleagues. Ineff is a friend of mine and he asked me today that very question; I thought about johnstone's sketches too. :) – tetrapharmakon May 9 2011 at 22:24
Following the previous indication of Professor Brown I want to add another possible way to see natural transformation which is a generalization of the previous definition.
Given categories $\mathcal C$ and $\mathcal D$ and two functors between them $\mathcal F,\mathcal G \colon \mathcal C \to \mathcal D$ then a natural transformation $\tau$ can be defined as a functor $\tau \colon \mathcal C \to (\mathcal F \downarrow \mathcal G)$ which arrow components are the diagonal functions, sending each arrow $f \in \mathcal C(c,c')$, with $c,c' \in \mathcal C$ to $(f,f) \in (\mathcal F \downarrow \mathcal G)(\tau(c),\tau(c'))$.
Edit: I think the definition of natural transformation proposed by professor Brown probably can be even a more natural than the one proposed in the question. I think that more details are worthed.
The key ingredient for that definition is the concept of arrow category of a given category $\mathbf D$: such category have morphism of $\mathbf D$ as objects and commutative square as morphisms.
This category come equipped with two functors $\mathbf {source}, \mathbf{target} \colon \text{Arr}(\mathbf D) \to \mathbf D$ such that for each object (i.e. a morphisms of $\mathbf D$) $f \colon d \to d'$ we have $$\mathbf{source}(f)=d$$ $$\mathbf{target}(f)=d'$$ while for each $f \in \mathbf D(x,x')$, $g \in \mathbf D(y,y')$ and a morphism $\alpha \in \text{Arr}(\mathbf D)(f,g)$ (i.e. a quadruple $\langle f,g, \alpha_0,\alpha_1\rangle$ where $\alpha_0 \in \mathbf D(x,y)$ and $\alpha_1 \in \mathbf D(x',y')$ such that $\alpha_1 \circ f = g \circ \alpha_0$) we have $$\mathbf{source}(\alpha)=\alpha_0$$ $$\mathbf{target}(\alpha)=\alpha_1$$ it's easy to prove that these data give two functors (which gives to $\text{Arr}(\mathbf D)$ the structure of a graph internal to $\mathbf{Cat}$).
Now let's take a look to this new definition of natural transformation:
A natural transformation $\tau$ between two functors $F,G \colon \mathbf C \to \mathbf D$ is a functor $\tau \colon \mathbf C \to \text{Arr}(\mathbf D)$ such that $\mathbf{source} \circ \tau = F$ and $\mathbf{target}\circ \tau = G$.
A functor of this kind associate to every object $c \in \mathbf C$ a morphism $\tau_c \colon F(c) \to G(c)$ in $\mathbf D$, while to every $f \in \mathbf C(c,c')$ it gives the commutative triangle expressing the equality $$\tau_{c'} \circ F(f)=\tau_{c'} \circ \mathbf {source}(\tau_f)=\mathbf {target}(\tau_f) \circ \tau_c = G(f) \circ \tau_c$$ certifying the naturality (in the ordinary sense) of the $\tau_c$. This definition reminds the notion of homotopy between maps $f,g \colon X \to Y$ as map of kind $X \to Y^I$ (i.e. an homotopy as a (continuous) family of path of $Y$).
That's not all, indeed we can reiterate the construction of the arrow category obtaining what I think is called a cubical set $$\mathbf D \leftarrow \text{Arr}(\mathbf D) \leftarrow \text{Arr}^2(\mathbf D)\leftarrow \dots$$ where each arrow should be thought as the pair of functors $\mathbf{source}{n+1},\mathbf{target}{n+1} \colon \text{Arr}^{n+1}(\mathbf D) \to \text{Arr}^n (\mathbf D)$.
In this way we can associate to each category a cubical set. There's also a natural way to associate to every functor a (degree 0) mapping of cubical sets.
If we consider natural transformation as maps from a category to an arrow category then this correspondence associate to each natural transformation a degree 1 map between such cubical sets (by degree one I mean that the induced map send every object of $\text{Arr}^n(\mathbf C)$ in an object of $\text{Arr}^{n+1}(\mathbf D)$). I've found really beautiful this construction because it shows an analogy between categories-functors-natural transformation and complexes-map of complexes-complexes homotopies.
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Charles Ehresmann had a natty way of developing natural transformations. For a category $C$ let $\square C$ be the double category of commuting squares in $C$. Then for a small category $B$ we can form Cat($B,\square_1 C$), the functors from $B$ to the direction 1 part of $\square C$. This gets a category structure from the category structure in direction 2 of $\square C$. So we get a category CAT($B,C$) of functors and natural transformations. This view makes it easier to verify the law
Cat($A \times B,C) \cong$Cat($A,$CAT($B,C$)).
And this method goes over to topological categories as well:
R. Brown and P. Nickolas, Exponential laws for topological categories, groupoids and groups and mapping spaces of colimits'', Cah. Top. G\'eom. Diff. 20 (1979) 179-198.
See also Section 6.5 of my book Topology and Groupoids for using the homotopy terminology for natural equivalences, as it was in the first 1968 edition entitled "Elements of Modern Topology" (McGraw Hill).
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http://unapologetic.wordpress.com/2008/01/02/connectedness/?like=1&source=post_flair&_wpnonce=a832ba55c5
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# The Unapologetic Mathematician
## Connectedness
Tied in with the fundamental notion of continuity for studying topology is the notion of connectedness. In fact, once two parts of a space are disconnected, there’s almost no topological influence of one on the other, which should be clear from an intuitive idea of what it might mean for a space to be connected or disconnected. This intuitive notion can be illustrated by considering subspaces of the real plane $\mathbb{R}^2$.
First, just so we’re clear, a subset of the plane is closed if it contains its boundary and open if it contains no boundary points. Here there’s a lot more between open and closed than there is for intervals with just two boundary points. Anyhow, you should be able to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same answer, but it’s interesting to generalize this to find a metric on the product of two arbitrary metric spaces.
Anyhow, back to connectedness. Take a sheet of paper to be your plane, and draw a bunch of blobs on it. Use dotted lines sometimes to say you’re leaving out that section of the blob’s border. Have fun with it.
Now we’ll consider that collection of blobs as a subspace $X\subseteq\mathbb{R}^2$, and thus it inherits the subspace topology. We can take one blob $B$ and throw an open set around it that doesn’t hit any other blobs (draw a dotted curve around the blob that doesn’t touch any other). Thus the blob is an open subset $B\subseteq X$ because it’s the intersection of the open set we drew in the plane and the subspace $X$. But we could also draw a solid curve instead of a dotted one and get a closed set in the plane whose intersection with $X$ is $B$. Thus $B$ is also a closed subset of $X$. Some people like to call such a subset in a topological space “clopen”.
In general, given any topological space we can break it into clopen sets. If the only clopen sets are the whole space and the empty subspace, then we’re done. Otherwise, given a nontrivial clopen subset, its complement must also be clopen (why?), and so we can break it apart into those pieces. We call a space with no nontrivial clopen sets “connected”, and a maximal connected subspace $B$ of a topological space $X$ we call a “connected component”. That is, if we add any other points from $X$ to $B$, it will be disconnected.
An important property of connected spaces is that we cannot divide them into two disjoint nonempty closed subsets. Indeed, if we could then the complement of one closed subset would be the other. It would be open (as the complement of a closed subset) and closed (by assumption) and nontrivial since neither it nor its complement could be empty. Thus we would have a nontrivial clopen subset, contrary to our assumptions.
If we have a bunch of connected spaces, we can take their coproduct — their disjoint union — to get a disconnected space with the original family of spaces as its connected components. Conversely, any space can be broken into its connected components and thus written as a coproduct of connected spaces. In general, morphisms from the coproduct of a collection of objects exactly correspond to collections of morphisms from the objects themselves. Here this tells us that a continuous function from any space is exactly determined by a collection of continuous functions, one for each connected component. So we don’t really lose much at all by just talking about connected spaces and trying to really understand them.
Sometimes we’re just looking near one point or another, like we’ve done for continuity or differentiability of functions on the real line. In this case we don’t really care whether the space is connected in general, but just that it looks like it’s connected near the point we care about. We say that a space is “locally connected” if every point has a connected neighborhood.
Sometimes just being connected isn’t quite strong enough. Take the plane again and mark axes. Then draw the graph of the function defined by $f(x)=\sin\left(\frac{1}{x}\right)$ on the interval $\left(0,1\right]$, and by $f(0)=0$. We call this the “topologist’s sine curve”. It’s connected because any open set we draw containing the wiggly sine bit gets the point $(0,0)$ too. The problem is, we might want to draw paths between points in the space, and we can’t do that here. For two points in the sine part, we just follow the curve, but we can never quite get to or away from the origin. Incidentally, it’s also not locally connected because any small ball around the origin contains a bunch of arcs from the sine part that aren’t connected to each other.
So when we want to draw paths, we ask that a space be “path-connected”. That is, given points $x_0$ and $x_1$ in our space $X$, there is a function $f:\left[0,1\right]\rightarrow X$ with $f(0)=x_0$ and $f(1)=x_1$. Slightly stronger, we might want to require that we can choose this function to be a homeomorphism from the closed interval $\left[0,1\right]$ onto its image in $X$. In this case we say that the space is “arc-connected”.
Arc-connectedness clearly implies path-connectedness, and we’ll see in more detail later that path-connectedness implies connectedness. However, the converses do not hold. The topologist’s sine curve gives a counterexample where connectedness doesn’t imply path-connectedness, and I’ll let you try to find a counterexample for the other converse.
Just like we had local connectedness, we say that a space is locally path- or arc-connected if every point has a neighborhood which is path- or arc-connected. We also have path-components and arc-components defined as for connected components. Unfortunately, we don’t have as nice a characterization as we did before for a space in terms of its path- or arc-components. In the topologist’s sine curve, for example, the wiggly bit and the point at the origin are the two path components, but they aren’t put together with anything so nice as a coproduct.
[UPDATE]: As discussed below in the comments, I made a mistake here, implicitly assuming the same thing edriv said explicitly. As I say below, point-set topology and analysis really live on the edge of validity, and there’s a cottage industry of crafting counterexamples to all sorts of theorems if you weaken the hypotheses just slightly.
### Like this:
Posted by John Armstrong | Point-Set Topology, Topology
## 12 Comments »
1. When you defined clopen sets, I immediately realized that they’re closed under finite intersection and union, but in fact they’re colesed under arbitrary intersection and union, arent’t they?
Comment by | January 2, 2008 | Reply
2. I don’t think that’s necessarily so, but I don’t have a counterexample offhand. Point-set topology (like analysis) has a way of breaking when you give it edge cases. There’s even a book (which might have this answer) called Counterexamples in Topology.
Comment by | January 2, 2008 | Reply
3. Oops in fact we have an easy counterexample: $w + 1$ (or $\{0\} \cup \{\frac 1n | n \in \mathbb{N+}\}$. I thought that the connected components were always clopen, but this is not the case (they’re just closed).
Comment by | January 2, 2008 | Reply
4. Two comments. First: one way to get interesting counterexamples to edriv’s first comment is via Stone duality, which (loosely) says that the category of Boolean algebras is equivalent to the opposite of so-called Stone spaces (compact, Hausdorff, totally disconnected spaces). Take a Boolean algebra A; there’s a way of cooking up a space Spec(A) whose points are Boolean algebra maps A –> 2 (the topology is the Zariski topology). In the other direction, given a Stone space X, we can look at continuous maps X –> 2 where 2 has the discrete topology (notice that these are tantamount to clopen sets of X). Clopen(X) forms a Boolean algebra. These two functors form part of a contravariant adjunction which is in fact a contravariant equivalence.
The point is that Clopen(X) can be any Boolean algebra; therefore we shouldn’t expect Clopen(X) to be necessarily complete (i.e., closed under unions and intersections). For a concrete example, take A to be the free Boolean algebra on the countable set N. By definition of “free”, the corresponding Stone space is the space of functions N –> 2, better known as Cantor space C. By Stone duality, we should be able to retrieve A as Clopen(C). We see from this example that the clopens of Cantor space are not closed under arbitrary unions or intersections (since A is not complete).
Second comment: it’s not true that every space is the coproduct of its connected components. An easy example is the space of rationals Q, which is totally disconnected (i.e., the connected components are just points). Come to think of it, just about any nontrivial Stone space would also provide a counterexample, by the same reasoning.
Comment by Todd Trimble | January 3, 2008 | Reply
5. Todd, I think you’re right.. I may have to go back and tweak a bit. In my head I think I might have been making edriv’s mistake, but not consciously. It’s a bit late, though, so I’ll try getting to that in the morning.
Comment by | January 3, 2008 | Reply
6. [...] of a Connected Space One theorem turns out to be very important when we’re dealing with connected spaces, or even just with a connected component of a space. If is a continuous map from a connected space [...]
Pingback by | January 3, 2008 | Reply
7. [...] Image of a Compact Space One of the nice things about connectedness is that it’s preserved under continuous maps. It turns out that compactness is the same way [...]
Pingback by | January 16, 2008 | Reply
8. [...] the hyperplanes for cannot fill up all of . What’s left over they chop into a bunch of connected components, which we call Weyl chambers. Thus every regular vector belongs to exactly one of these Weyl [...]
Pingback by | February 3, 2010 | Reply
9. [...] — the identified base for the topology on — consists of all the subsets of which are clopen — both open and closed. That is, elements of correspond to unions of connected components of [...]
Pingback by | August 19, 2010 | Reply
10. [...] only way this is possible is for and to live in different connected components of . And, indeed, our definition so far doesn’t rule out this possibility. We could have a [...]
Pingback by | February 24, 2011 | Reply
11. [...] with such a sequence of parallelepipeds is open, as is the set of points we can’t; since is connected, only one of these can be nonempty, and since we can surely reach any point in , the set of points [...]
Pingback by | December 6, 2011 | Reply
12. [...] let’s define a function. In every connected component of , pick some base-point . As an aside, what we really want are the arc components of , but since [...]
Pingback by | December 15, 2011 | Reply
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http://math.stackexchange.com/questions/272057/maximal-dimension-of-subspace-of-matrices-whose-products-have-zero-trace
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Maximal dimension of subspace of matrices whose products have zero trace
In the space of all real matrices with dimension $n$, find the maximal possible dimension of a subspace $V$ such that $\forall X,Y\in V,\, \operatorname{tr}(XY)=0$.
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2 Answers
If $V$ is a subspace of real matrices of size $n$ with the property that $\operatorname{tr}(XY)=0$ for all $X,Y\in V$, then $\operatorname{tr}(X^2)=0$ for all $X\in V$.
Let $W$ be a subspace of real matrices of size $n$ with the property that $\operatorname{tr}(X^2)=0$ for all $X\in W$. Then $\operatorname{tr}((X+Y)^2)=0$ for all $X,Y\in W$. But $\operatorname{tr}((X+Y)^2)=\operatorname{tr}(X^2)+\operatorname{tr}(Y^2)+2\operatorname{tr}(XY)$. It follows that $\operatorname{tr}(XY)=0$ for all $X,Y\in W$.
This shows that the problem reduces to the following: Investigations about the trace form which is already solved.
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its all very nice! – dineshdileep Jan 8 at 8:10
(This is not a difficult problem, but it is also not an easy homework problem. I wonder if the original problem is much simpler and the OP has copied the problem statement wrongly.)
Note that $\operatorname{tr}(XY)=\operatorname{vec}(Y^T)^T\operatorname{vec}(X)= \operatorname{vec}(Y)^TP\operatorname{vec}(X)$, where $P$ is the symmetric permutation matrix defined by $P_{(j-1)n+i,\ (i-1)n+j}=P_{(i-1)n+j,\ (j-1)n+i}=1$ for $i,j\in\{1,\ldots,n\}$ and other entries zero. Since the matrix $P$ is orthogonally similar to $I_p\oplus (-I_q)$, where $q=n(n-1)/2$ and $p=n^2-q$, a maximal matrix subspace $V$ such that $\operatorname{tr}(XY)=0$ whenever $X,Y\in V$ can be identified with a maximal subspace $W\subset\mathbb{R}^{n^2}$ such that $u^T(I_p\oplus(-I_q))v=0$ whenever $u,v\in W$.
We now claim that $\dim W\le q$. Suppose the contrary. Then there exists a set of $q+1$ linearly independent vectors $u_1,\ldots,u_{q+1}\in\mathbb{R}^{n^2}$ such that $u_i^T(I_p\oplus(-I_q))u_j=0$ for all $i,j=1,\ldots,q+1$. Hence there exists a nonzero linear combination $x=\sum_i c_iu_i\not=0$ such that the last $q$ entries of $x$ are zero. Write $x^T=(\tilde{x}^T,0_{1\times q})$. Then $x^T(I_p\oplus(-I_q))x=\|\tilde{x}\|^2\not=0$ because $\tilde{x}\not=0$, but we also have $x^T(I_p\oplus(-I_q))x=0$ because $x$ is a linear combination of $u_1,\ldots,u_{q+1}$. So we arrive at a contradiction. Therefore $\dim W\le q$.
Finally, the dimension $q$ is attainable. For example, consider the set of all strictly upper triangular matrices. Hence the required maximal dimension is $q=n(n-1)/2$.
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@YACP The OP has asked several problems in last few hours, and all but this one are typical homework problems. – user1551 Jan 7 at 16:19
seems very elegant – dineshdileep Jan 8 at 8:52
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http://ams.org/bookstore?fn=20&arg1=survseries&ikey=SURV-89.S
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The Concentration of Measure Phenomenon
Michel Ledoux, Université Paul-Sabatier, Toulouse, France
SEARCH THIS BOOK:
Mathematical Surveys and Monographs
2001; 181 pp; softcover
Volume: 89
Reprint/Revision History:
reprinted 2005
ISBN-10: 0-8218-3792-3
ISBN-13: 978-0-8218-3792-4
List Price: US\$68
Member Price: US\$54.40
Order Code: SURV/89.S
It was undoubtedly a necessary task to collect all the results on the concentration of measure during the past years in a monograph. The author did this very successfully and the book is an important contribution to the topic. It will surely influence further research in this area considerably. The book is very well written, and it was a great pleasure for the reviewer to read it.
--Mathematical Reviews
The observation of the concentration of measure phenomenon is inspired by isoperimetric inequalities. A familiar example is the way the uniform measure on the standard sphere $$S^n$$ becomes concentrated around the equator as the dimension gets large. This property may be interpreted in terms of functions on the sphere with small oscillations, an idea going back to Lévy. The phenomenon also occurs in probability, as a version of the law of large numbers, due to Emile Borel. This book offers the basic techniques and examples of the concentration of measure phenomenon. The concentration of measure phenomenon was put forward in the early seventies by V. Milman in the asymptotic geometry of Banach spaces. It is of powerful interest in applications in various areas, such as geometry, functional analysis and infinite-dimensional integration, discrete mathematics and complexity theory, and probability theory. Particular emphasis is on geometric, functional, and probabilistic tools to reach and describe measure concentration in a number of settings.
The book presents concentration functions and inequalities, isoperimetric and functional examples, spectrum and topological applications, product measures, entropic and transportation methods, as well as aspects of M. Talagrand's deep investigation of concentration in product spaces and its application in discrete mathematics and probability theory, supremum of Gaussian and empirical processes, spin glass, random matrices, etc. Prerequisites are a basic background in measure theory, functional analysis, and probability theory.
Readership
Graduate students and research mathematicians interested in measure and integration, functional analysis, convex and discrete geometry, and probability theory and stochastic processes.
• Concentration functions and inequalities
• Isoperimetric and functional examples
• Concentration and geometry
• Concentration in product spaces
• Entropy and concentration
• Transportation cost inequalities
• Sharp bounds of Gaussian and empirical processes
• Selected applications
• References
• Index
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http://www.physicsforums.com/showthread.php?t=141301
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Physics Forums
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## The Massless Photon
(I hope I've chosen the right sub-forum for this question...)
Hi folks - I've recently joined here to see if people who are more knowledgeable than me can help me understand some physics issues I have struggled with for a long time.
My current question is a pretty basic one about how it is possible for a photon to have no mass.
We have the famous equation, "E = mc squared" My math knowledge is very limited, but from what I know - if I assign the value "0" to m, and multiply 0 by c squared, the answer for E should be zero.
Yet a photon possesses energy, and is said to have no mass.
I can see 3 possibilities:
- "E = mc squared" is not a standard algebra equation, and assigning the value "0" to m doesn't result in E being zero.
- "E = mc squared" does not apply to photons - something that seems very unlikely to me.
- photons do, in fact, have mass, or conversely, have no energy.
Can someone help me understand this? I would be very grateful. Thanks!
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
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It's usually a good idea to start out on wikipedia before asking here. http://en.wikipedia.org/wiki/E%3Dmc%C2%B2
## The Massless Photon
Quote by Thrice It's usually a good idea to start out on wikipedia before asking here. http://en.wikipedia.org/wiki/E%3Dmc%C2%B2
Actually I do usually start with Wiki, but it never occurred to me to look up "E = mc squared" there.
Thanks for the link - it answers my question.
That was easy, wasn't it? Carry on, carry on...
Quote by Lelan Thara Yet a photon possesses energy, and is said to have no mass. Can someone help me understand this? I would be very grateful. Thanks!
If you’re just looking for a way to understand; there is a very simple way.
Don't thing of a photon as "possessing" energy --- thing of a photon AS energy.
Now when an atom absorbs a photon and moves one its electrons into a higher energy ‘obit’ it gains mass.
When it drops to a lower energy level the atom loses mass as a photon is emitted in some random direction.
Where was the mass? in the electron; or in the atom as whole – I’ll let you speculate.
But that mass cannot just appear and disappear; shouldn’t mass be “conserved”?
No conservation of mass and conservation of energy is “Old School”
In modern physics it is the net of Mass and Energy that is conserved.
So when a bit of mass disappears from an atom we can find it in the energy that escapes from it – we call it a photon.
When a lot of mass disappears very quickly we can describe it as a lot photons (aka energy) escaping – or more easily described as a nuclear explosion.
Mentor
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Quote by Lelan Thara My current question is a pretty basic one about how it is possible for a photon to have no mass. We have the famous equation, "E = mc squared" My math knowledge is very limited, but from what I know - if I assign the value "0" to m, and multiply 0 by c squared, the answer for E should be zero. Yet a photon possesses energy, and is said to have no mass.
The complete relativistic expression for the energy of a particle is this:
$$E = \sqrt{p^2c^2 + m^2c^4}$$
Where p is the momentum. Note that for massive particles at rest (momentum = 0) that equation becomes the more familiar $E = mc^2$.
For a photon: the mass is zero, but the momentum and energy are non-zero. E = pc = hf.
Quote by Lelan Thara Actually I do usually start with Wiki, but it never occurred to me to look up "E = mc squared" there. Thanks for the link - it answers my question. That was easy, wasn't it? Carry on, carry on...
Glad I could help. I picked up this firefox extension to make sure I can't avoid searching on wikipedia.
https://addons.mozilla.org/firefox/2517/
Thanks agains to all. The most fundamental answer to my question is that E=mc squared is not as universally applied as I assumed it was, at least not in its familiar simple form. There is something that still confuses me a bit, and that's the concept of "rest mass". I was under the impression that the only time mass could be at rest was at a temperature of absolute zero, which doesn't really exist in nature (like a perfect vacuum). The photon is also never at rest - is the difference that the photon can't conceptually be at rest? I've also come across references to the "virtual mass" of a photon. Can anyone shed more light on that? The larger question might be - is the massless photon massless by definition? In other words, "mass is a quality of matter, the photon isn't defined as matter despite being a particle, and therefore the photon can't have a quality of matter - mass"? Or is it something more concrete - that the photon doesn't exhibit inertia, acceleration and deceleration, and other properties of mass? Thanks again.
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Quote by Lelan Thara Thanks agains to all. The most fundamental answer to my question is that E=mc squared is not as universally applied as I assumed it was, at least not in its familiar simple form. There is something that still confuses me a bit, and that's the concept of "rest mass". I was under the impression that the only time mass could be at rest was at a temperature of absolute zero, which doesn't really exist in nature (like a perfect vacuum). The photon is also never at rest - is the difference that the photon can't conceptually be at rest?
That's in quantum mechanics. Special Relativity CAN be interfaced to quantum mechanics, but when we talk about SR by itself we mostly mean the classical model, where things can be at rest. You are right about the photon though, it has no rest frame, even in the classical theory, and no mass.
Sidebar.
I am one of the ones who do not use "rest mass" because that implies there is some kind of distinguished non-rest mass, which I deny. There are are other posters on this forum who vehemently disagree, and in order to get maximal use out of your experience here you have to be clear in your mind so the disagreement (which is a tempest in a teacup as far as physics is concerned) doesn't confuse you. In my view a particle has one mass, no matter how it is moving in relation to you, and this mass is a "scalar", that is to say the same in every frame.
I've also come across references to the "virtual mass" of a photon. Can anyone shed more light on that?
I don't know this one. Did you mean virtual photons?
The larger question might be - is the massless photon massless by definition? In other words, "mass is a quality of matter, the photon isn't defined as matter despite being a particle, and therefore the photon can't have a quality of matter - mass"?
ST postulates that the speed of light is invariant, the same for all inertial frames. It is a mathematical deduction from that postulate that if light is carried by a particle then that particle must be massless. In fact the two concepts "Massless" and "Moves at the Speed of Light" are synonymous in SR. Gluons in the Standard Model also have this property.
Or is it something more concrete - that the photon doesn't exhibit inertia, acceleration and deceleration, and other properties of mass?
As far as we can tell, it takes no acceleration to get a massless particle up to the speed c. But the experimental evidence for the masslessness of the photon is mostly in another area. From classical wave theory it follows that if light were carried by a massive particle, it would exhibit longitudinal (compression type) waves. Whereas we only find transverse light waves. Experimentalists do tests with fantastically refined versions of this distinction to examine the possibility that there are some very weak compression waves, that would allow a tiny mass to the photon. They've been doing this, with ever increasing precision, for generations. They've never found anything.
Of course no experiment can ever proves that the mass is exactly zero; every experiment has error bounds, but the error bounds on these tests have gotten really tiny, off the top of my head I want to say + or - 10-20 electron volts.
Thanks again.[/QUOTE]
Quote by Lelan Thara Thanks agains to all. The most fundamental answer to my question is that E=mc squared is not as universally applied as I assumed it was, at least not in its familiar simple form. There is something that still confuses me a bit, and that's the concept of "rest mass". I was under the impression that the only time mass could be at rest was at a temperature of absolute zero, which doesn't really exist in nature (like a perfect vacuum). The photon is also never at rest - is the difference that the photon can't conceptually be at rest? I've also come across references to the "virtual mass" of a photon. Can anyone shed more light on that? The larger question might be - is the massless photon massless by definition? In other words, "mass is a quality of matter, the photon isn't defined as matter despite being a particle, and therefore the photon can't have a quality of matter - mass"? Or is it something more concrete - that the photon doesn't exhibit inertia, acceleration and deceleration, and other properties of mass? Thanks again.
You made a good point here. The concept of a "massless particle" is so familiar in present day physics that people forget that it is in fact not such a good concept at all. As you pointed out, the photon is indeed never at rest, and can not be, so why should one talk of rest mass ? It is really nonsense if you think about it and it seems to cause a lot of confusion to many people (judging from the forum questions).So, the photon is massless by definition only with the convenient result that one can insert m=0 into the complete relativistic expression, such that one obtains the correct relation between energy and impulse. Maybe it would be better to use the term "c-particle" to indicate that it is a particle which moves at the speed of light.
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Quote by notknowing Maybe it would be better to use the term "c-particle" to indicate that it is a particle which moves at the speed of light.
There already is a term: "lightlike" particle.
Quote by Lelan Thara Thanks agains to all. The most fundamental answer to my question is that E=mc squared is not as universally applied as I assumed it was, at least not in its familiar simple form.
it could be more universally applied if the $m$ in $E = m c^2$ is always considered the relativistic mass and $E$ is the total energy, kinetic energy plus rest energy. or the energy of the particle in the frame of the observer watching it whiz by. or the momentum of the particle divided by speed. assuming photons travel at the speed of the wavespeed of electromagnetic radiation, $c$, the (relativistic) mass of the photons is
$$m = \frac{E}{c^2} = \frac{h \nu}{c^2}$$
but here $m$ is not the "rest mass" or "invariant mass". for particles that move more slowly that $c$, special relativity says their momentum is
$$p = m v = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$
where $m_0$ is the rest mass (and $E_0 = m_0 c^2$ is the rest energy and the total energy $E_0 = m_0 c^2$ is the rest energy plus kinetic energy). so the relationship between rest mass and relativistic mass is
$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$
or turned around is
$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$
or
$$m_0 = \frac{E}{c^2} \sqrt{1 - \frac{v^2}{c^2}}$$ .
now, no matter what finite energy that particle has, if it moves at the speed of light, that equation says that the rest mass of the particle has to be zero.
the equation
$$E = m c^2$$
is just as general as
$$E^2 = m_0^2 c^4 + p^2 c^2$$
if the rest mass $m_0$ is related to relativistic mass $m$ as per the equation above and momentum is stiil the same $p = m v$.
There is something that still confuses me a bit, and that's the concept of "rest mass".
particles moving past an observer at very high speeds "appear" to that observer to have a larger mass than they do if the observer is moving alongside the particle. the mass that the particle has in the same reference frame of the particle is the rest mass.
I've also come across references to the "virtual mass" of a photon. Can anyone shed more light on that?
dunno if it means
$$m = \frac{h \nu}{c^2}$$
of if it means that the jury might be out as to whether photons, the particle manifestation of light, travels as fast as the wavespeed of light. some have posted is upper bound for the rest masses of photons (if i recall this upper bound was somewhere around 10-52 kg which is virtually nothing. so the difference in speeds are not measureable if there is such a difference at all. if it turns out that photons are known to travel at precisely the speed of light (waves), then the rest mass of the photons would have to be zero.
The larger question might be - is the massless photon massless by definition? In other words, "mass is a quality of matter, the photon isn't defined as matter despite being a particle, and therefore the photon can't have a quality of matter - mass"?
it's because they (are believed to) move at the speed of light that their rest mass has to be zero.
Or is it something more concrete - that the photon doesn't exhibit inertia, acceleration and deceleration, and other properties of mass?
it exhibits inertia, but no acceleration or deceleration if it always flies by at a speed of $c$, no matter who the observer is (this is a postulate of special relativity). they have momentum of
$$p = m v = \frac{h \nu}{c^2} v$$
which, if they move at speed $v = c$, then the momentum is
$$p = \frac{h \nu}{c}$$
they have non-zero momentum if that is what you mean by "inertia".
i know that i am presenting this from a POV that is discouraged in modern physic pedagogy (i don't think that Doc Al will like it), but it's correct given the definition of relativistic mass.
Thank you, folks. I wish I could tell you where I saw the "virtual mass" for a photon mentioned. Most of my physics reading is books for laymen, but once I did try to wade through articles by physicsts themselves in the library - I think it might have be in an article by Bohr where I saw it. Here's a related question - if a photon has no mass, what pushes a solar sail? Is it other forms of radiation, as opposed to photons?
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Quote by Lelan Thara Here's a related question - if a photon has no mass, what pushes a solar sail? Is it other forms of radiation, as opposed to photons?
Photons have momentum, as well as energy.
Quote by Lelan Thara (I hope I've chosen the right sub-forum for this question...) Hi folks - I've recently joined here to see if people who are more knowledgeable than me can help me understand some physics issues I have struggled with for a long time. My current question is a pretty basic one about how it is possible for a photon to have no mass. We have the famous equation, "E = mc squared" My math knowledge is very limited, but from what I know - if I assign the value "0" to m, and multiply 0 by c squared, the answer for E should be zero. Yet a photon possesses energy, and is said to have no mass. I can see 3 possibilities: - "E = mc squared" is not a standard algebra equation, and assigning the value "0" to m doesn't result in E being zero. - "E = mc squared" does not apply to photons - something that seems very unlikely to me. - photons do, in fact, have mass, or conversely, have no energy. Can someone help me understand this? I would be very grateful. Thanks!
The answer to your question regards your confusion as to the definition of the term "mass." Sometimes the term is used to refer to a particles proper mass while sometimes its used to refer to inertial mass (aka relativistic mass). The photon has zero proper mass and an inertial mass m = E/c2. For details please see
http://www.geocities.com/physics_world/mass_paper.pdf
Best wishes
Pete
Recognitions: Science Advisor Staff Emeritus Note that Pete's paper, with the URL given above, while mostly correct, has not been peer reviewed, and that some of us (like me) have disagreements with him on certain technical points and usages. Most of these are rather "fine" points, though.
Pervect - given what you said above, my question would be - do you agree with Pete that photons have inertial mass?
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http://math.stackexchange.com/questions/40564/what-is-the-symmetry-between-the-definitions-of-the-bounded-universal-existentia/40570
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# What is the symmetry between the definitions of the bounded universal/existential quantifiers?
What is the symmetry between the definitions of the bounded universal/existential quantifiers?
$\forall x \in A, B(x)$ means $\forall x (x \in A \rightarrow B(x))$
$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$
These make intuitive sense, but I would expect there to be some kind of symmetry between how the definitions of the bounded quantifiers work, and I can't see one. $A \rightarrow B$ means $\lnot A \lor B$ which doesn't seem to have a direct relationship with $A \land B$. What am I missing?
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I am not sure what you mean by symmetry, but the existencial and universal quantifiers are respectively, the left and right adjoints of the inverse image functor. – G. Rodrigues May 23 '11 at 17:29
It has been quite some time since you have asked the question, is there anything missing from the answers? If so, please update the question accordingly; otherwise please accept one of the answers. – Asaf Karagila Jul 13 '11 at 9:41
I haven't been able to clarify this for myself. I will review the answers and see if I can understand what I still don't get tomorrow. – John Salvatier Jul 15 '11 at 16:14
## 4 Answers
Consider $\forall x\in A, B(x)$. That is if $x$ is a member of $A$ then $B(x)$ is true.
This translates, as you well put to $\forall x(x\in A\rightarrow B(x))$, which again translates to $\forall x(x\notin A\lor B(x))$.
Denote $B=\{x\mid B(x)\}$, that is all the elements which satisfy $B(x)$. Now you have that $x\in B \iff B(x)$ is true.
We re-evaluate the previous sentence: $\forall x(x\in A\rightarrow x\in B)$, that is to say $A\subseteq B$, or to say $\forall x(x\notin A\lor x\in B)$.
Now consider the existential version: $\exists x(x\in A\land B(x))$ that is $\exists x(x\in A\land x\in B)$, or in simpler words $A\cap B\neq\emptyset$.
Lastly we recall that $\lnot\forall x\varphi(x) \iff \exists x\lnot\varphi(x)$. This translates well over bounded quantifiers:
$$\begin{align} \lnot(\forall x\in A, B(x)) &\iff \lnot\forall x(x\in A\rightarrow x\in B)\\ &\iff \exists x\lnot(x\in A\rightarrow x\in B)\\ &\iff \exists x(x\in A\land x\notin B)\\ &\iff \exists x\in A,\lnot B(x) \end{align}$$
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I think this shows that Exists and Forall are simply less symmetric than I was expecting. – John Salvatier Jul 17 '11 at 1:28
@John: Could you elaborate? – Asaf Karagila Jul 17 '11 at 5:21
You might think of universal as a mega-intersection, and existential as a mega-union. E.g., if $A=\lbrace x_1,x_2,\dots\rbrace$ then your first formula is $B(x_1)$ and $B(x_2)$ and ..., while the second is $B(x_1)$ or $B(x_2)$ or ....
There is also some sort of symmetry in noting that "not for all" is the same as "there exists ... not," and "not there exists" is the same as "for all ... not."
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I don't think this answers the question but +1 for the notion of forall being an extended conjunction and exists being an extended disjunction. I hadn't heard that notion before. Do you know of something that explores this notion? – John Salvatier May 22 '11 at 6:52
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@John: It is a standard argument, in fact if you look closely you realize that it is actually the definition. In older papers $\forall$ was denoted by $\bigwedge$ and $\exists$ was denoted by $\bigvee$. – Asaf Karagila May 22 '11 at 8:39
Is this more symmetrical?
$\forall x \in A, B(x)$ means $\forall x (x \notin A \lor B(x))$
$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$
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How is that different from what I wrote? – Asaf Karagila May 22 '11 at 10:37
You mention the above formula in only in passing. – Dietrich Epp May 22 '11 at 10:45
As Asaf Karagila says, those two connectives, $\forall x \in A, \ldots$ and $\exists x \in A, \ldots$ are dual of each other.
Just like the conjunction / disjunction pair or the universal / existential pair, we have that $\lnot (\forall x \in A, \phi(x)) \Leftrightarrow \exists x \in A, \lnot \phi(x)$.
Also, if $A$ is a singleton $\{a\}$, those two connective are logically equivalent, and $\forall x \in A, \phi(x) \Leftrightarrow \exists x \in A, \phi(x) \Leftrightarrow \phi(a)$
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http://math.stackexchange.com/questions/4844/distributions-of-point-charges/5326
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Distributions of point charges
Problem
$N$ point charges are distributed in the unit ball in $\mathbb{R}^k$, $k=2,3$. Given locations of the particles $x_1,\ldots,x_N$ the potential energy is
$E=\sum_{j=1}^{N-1}\sum_{k=j+1}^N |x_j-x_k|^{-1}$
where $|x_j-x_k|$ is Euclidean distance between $x_j$ and $x_k$. I'm interested in both the minimal value of $E$ over all possible locations of the particles in the unit ball and what this configuration looks like.
On the Unit Interval For $k=1$ the $N$ charges are distributed on the interval $[-1,1]$ according to the roots of the $N+1$th Chebshev polynomial. See: http://en.wikipedia.org/wiki/Chebyshev_polynomials#Roots_and_extrema
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5
I can do no better than point out dx.doi.org/10.1088/0305-4470/31/3/014 and dx.doi.org/10.1098/rspa.2001.0913 . – J. M. Sep 17 '10 at 9:36
1
@J.M. Thank you very much! – alext87 Sep 17 '10 at 14:39
1
This problem was much harder than I indeed. As J.M. kindly pointed out in his comment it appears that for $k=2$ the problem has been approximately solved for $N<80$ and for $k=3$ only for $N<32$. – alext87 Sep 17 '10 at 18:45
Can't you do this with a (kinetic) Monte Carlo algorithm ? – max Jul 18 '11 at 21:08
4 Answers
The canonical thing to do for a question like this is to look at Neal Sloane's home page. Sure enough, there is a table giving some good arrangements.
http://www2.research.att.com/~njas/electrons/index.html
This was indeed one of the links on the page in wok's answer, but it may be the most complete resource.
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You can look at one particular configuration at "Animated (Java) Illustrations {of} 24 Electrons on a Sphere" and a few more at "Min-Energy Configurations of Electrons On A Sphere".
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Your first link is broken; it ends with ".ht" where it should be ".htm". – Rahul Narain Sep 24 '10 at 22:07
@Rahul : Thank you. fixed. – David Cary Sep 27 '10 at 22:03
The most comprehensive webpage I have found is about evenly distributing N points on a sphere. To be more general, this is known as the seventh of Stephen Smale's problems: the "optimal" distribution of points on the 2-sphere. It is still unsolved.
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This problem seems related to evenly distributing points across the surface of a sphere (specifically, that is very similar to your k=2 case). That problem is addressed in a programming competition challenge named PSPHERE at SPOJ. The solutions there are not public, but perhaps approaching the leading contestants on that particular challenge could prove helpful.
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http://mathhelpforum.com/new-users/198145-trying-remember-maths-after-40-years.html
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# Thread:
1. ## Trying to remember maths after 40 years
I have forgotten some/ most / all of my maths knowledge.
I am trying to solve
x=10-x2 *t
for x with respect to t.
I know that I used to be able to do this and I also know that I thought it was simple but my memory has gone and who keeps text books for 40 years.
Thanks for any help
2. ## Re: Trying to remember maths after 40 years
Originally Posted by KRe
I have forgotten some/ most / all of my maths knowledge.
I am trying to solve
x=10-x2 *t
for x with respect to t.
I know that I used to be able to do this and I also know that I thought it was simple but my memory has gone and who keeps text books for 40 years.
Thanks for any help
$\displaystyle \begin{align*} x &= 10 - t\,x^2 \\ t\,x^2 + x &= 10 \\ x^2 + \frac{1}{t}\,x &= \frac{10}{t} \\ x^2 + \frac{1}{t}\,x + \left(\frac{1}{2t}\right)^2 &= \frac{10}{t} + \left(\frac{1}{2t}\right)^2 \\ \left(x + \frac{1}{2t}\right)^2 &= \frac{40t}{4t^2} + \frac{1}{4t^2} \\ \left(x + \frac{1}{2t}\right)^2 &= \frac{40t + 1}{4t^2} \\ x + \frac{1}{2t} &= \pm \frac{\sqrt{40t + 1}}{2t} \\ x &= -\frac{1}{2t} \pm \frac{\sqrt{40t+1}}{2t} \\ x = \frac{-1 - \sqrt{40t + 1}}{2t} \textrm{ or } x &= \frac{-1 + \sqrt{40t + 1}}{2t} \end{align*}$
3. ## Re: Trying to remember maths after 40 years
If you knew algebra once and you want to bring the memory back then look for a copy of "Forgotten Algebra." It is an inexpensive paperback written for exactly this purpose and is available used or in libraries. Do every problem in the book as fast as you can go and then do them all over again. You should find your memory coming back after that.
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http://en.m.wikibooks.org/wiki/General_Relativity/Reissner-Nordstr%C3%B6m_black_hole
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# General Relativity/Reissner-Nordström black hole
<General Relativity
Reissner-Nordström black hole is a black hole that carries electric charge $Q$, no angular momentum, and mass $M$. General properties of such a black hole are described in the article charged black hole.
It is described by the electric field of a point-like charged particle, and especially by the Reissner-Nordström metric that generalizes the Schwarzschild metric of an electrically neutral black hole:
$ds^2=-\left(1-\frac{2M}{r}+\frac{Q^2}{r^2}\right)dt^2 + \left(1-\frac{2M}{r}+\frac{Q^2}{r^2}\right)^{-1} dr^2 +r^2 d\Omega^2$
where we have used units with the speed of light and the gravitational constant equal to one ($c=G=1$) and where the angular part of the metric is
$d\Omega^2 \equiv d\theta^2 +\sin^2 \theta\cdot d\phi^2$
The electromagnetic potential is
$A = -\frac{Q}{r}dt$.
While the charged black holes with $|Q| < M$ (especially with $|Q| << M$) are similar to the Schwarzschild black hole, they have two horizons: the event horizon and an internal Cauchy horizon. The horizons are located at $r = r_\pm := M \pm \sqrt{M^2-Q^2}$. These horizons merge for $|Q|=M$ which is the case of an extremal black hole.
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http://mathhelpforum.com/algebra/143144-algebraic-word-problems.html
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# Thread:
1. ## algebraic word problems
1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed?
2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?
3)two cars are traveling toward each other, starting at the same time 480km apart. if they meet after 4 hours, and one car is traveling 40kph faster than the other, how fast is each car moving?
4) bill has a collection of 65 nickels and dimes. their total worth is \$5.50. find how many nickels and dimes bill has.
can someone please show me how to do these??????
2. Originally Posted by pspman354
2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?
$\frac{x}{y}= \frac{6}{5}$ ...(1)
$x-y = 8$ ...(2)
Now solve.
3. Originally Posted by pspman354
1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed?
20 Litres of 65% acid solution = 20 x 0.65 = 13L acid
a Litres of 60% acid solution = 0.6a L acid
b Litres of 80% acid solution = 0.8b L acid.
You want
0.6a + 0.8b = 13 Litres of acid
a + b = 20 L solution
Solve for a and b.
Attempt the first one. Then apply the same method to the other questions.
4. $\frac{x}{y}= \frac{6}{5}$ ...(1)
how do i solve for x or y using this?
5. Originally Posted by pspman354
$\frac{x}{y}= \frac{6}{5}$ ...(1)
how do i solve for x or y using this?
you can't you need both equations
$\frac{x}{y}= \frac{6}{5}$ ...(1)
$x-y = 8$ ...(2)
Using (2) $x=8+y$ and back into (1) gives
$\frac{8+y}{y}= \frac{6}{5}$
now finish it.
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http://terrytao.wordpress.com/category/teaching/254a-random-matrices/page/2/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Category Archive
You are currently browsing the category archive for the ‘254A – random matrices’ category.
## 254A, Notes 1: Concentration of measure
3 January, 2010 in 254A - random matrices, math.PR | Tags: Azuma's inequality, Chernoff inequality, concentration of measure, McDiarmid's inequality, moment method, Talagrand inequality | by Terence Tao | 77 comments
Suppose we have a large number of scalar random variables ${X_1,\ldots,X_n}$, which each have bounded size on average (e.g. their mean and variance could be ${O(1)}$). What can one then say about their sum ${S_n := X_1+\ldots+X_n}$? If each individual summand ${X_i}$ varies in an interval of size ${O(1)}$, then their sum of course varies in an interval of size ${O(n)}$. However, a remarkable phenomenon, known as concentration of measure, asserts that assuming a sufficient amount of independence between the component variables ${X_1,\ldots,X_n}$, this sum sharply concentrates in a much narrower range, typically in an interval of size ${O(\sqrt{n})}$. This phenomenon is quantified by a variety of large deviation inequalities that give upper bounds (often exponential in nature) on the probability that such a combined random variable deviates significantly from its mean. The same phenomenon applies not only to linear expressions such as ${S_n = X_1+\ldots+X_n}$, but more generally to nonlinear combinations ${F(X_1,\ldots,X_n)}$ of such variables, provided that the nonlinear function ${F}$ is sufficiently regular (in particular, if it is Lipschitz, either separately in each variable, or jointly in all variables).
The basic intuition here is that it is difficult for a large number of independent variables ${X_1,\ldots,X_n}$ to “work together” to simultaneously pull a sum ${X_1+\ldots+X_n}$ or a more general combination ${F(X_1,\ldots,X_n)}$ too far away from its mean. Independence here is the key; concentration of measure results typically fail if the ${X_i}$ are too highly correlated with each other.
There are many applications of the concentration of measure phenomenon, but we will focus on a specific application which is useful in the random matrix theory topics we will be studying, namely on controlling the behaviour of random ${n}$-dimensional vectors with independent components, and in particular on the distance between such random vectors and a given subspace.
Once one has a sufficient amount of independence, the concentration of measure tends to be sub-gaussian in nature; thus the probability that one is at least ${\lambda}$ standard deviations from the mean tends to drop off like ${C \exp(-c\lambda^2)}$ for some ${C,c > 0}$. In particular, one is ${O( \log^{1/2} n )}$ standard deviations from the mean with high probability, and ${O( \log^{1/2+\epsilon} n)}$ standard deviations from the mean with overwhelming probability. Indeed, concentration of measure is our primary tool for ensuring that various events hold with overwhelming probability (other moment methods can give high probability, but have difficulty ensuring overwhelming probability).
This is only a brief introduction to the concentration of measure phenomenon. A systematic study of this topic can be found in this book by Ledoux.
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## 254A, Notes 0a: Stirling’s formula
2 January, 2010 in 254A - random matrices, math.CA | Tags: factorial function, gamma function, Stirling's formula, trapezoid rule | by Terence Tao | 25 comments
In this supplemental set of notes we derive some approximations for ${n!}$, when ${n}$ is large, and in particular Stirling’s formula. This formula (and related formulae for binomial coefficients ${\binom{n}{m}}$ will be useful for estimating a number of combinatorial quantities in this course, and also in allowing one to analyse discrete random walks accurately.
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## 254A, Notes 0: A review of probability theory
1 January, 2010 in 254A - random matrices, math.CA, math.PR | Tags: almost sure convergence, conditioning, convergence in probability, independence, moments, random variables | by Terence Tao | 93 comments
In preparation for my upcoming course on random matrices, I am briefly reviewing some relevant foundational aspects of probability theory, as well as setting up basic probabilistic notation that we will be using in later posts. This is quite basic material for a graduate course, and somewhat pedantic in nature, but given how heavily we will be relying on probability theory in this course, it seemed appropriate to take some time to go through these issues carefully.
We will certainly not attempt to cover all aspects of probability theory in this review. Aside from the utter foundations, we will be focusing primarily on those probabilistic concepts and operations that are useful for bounding the distribution of random variables, and on ensuring convergence of such variables as one sends a parameter ${n}$ off to infinity.
We will assume familiarity with the foundations of measure theory; see for instance these earlier lecture notes of mine for a quick review of that topic. This is also not intended to be a first introduction to probability theory, but is instead a revisiting of these topics from a graduate-level perspective (and in particular, after one has understood the foundations of measure theory). Indeed, I suspect it will be almost impossible to follow this course without already having a firm grasp of undergraduate probability theory.
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## Course announcement: 254A – random matrices
3 December, 2009 in 254A - random matrices, math.PR, math.SP | Tags: random matrices | by Terence Tao | 8 comments
Starting in the winter quarter (Monday Jan 4, to be precise), I will be giving a graduate course on random matrices, with lecture notes to be posted on this blog. The topics I have in mind are somewhat fluid, but my initial plan is to cover a large fraction of the following:
• Central limit theorem, random walks, concentration of measure
• The semicircular and Marcenko-Pastur laws for bulk distribution
• A little bit on the connections with free probability
• The spectral distribution of GUE and gaussian random matrices; theory of determinantal processes
• A little bit on the connections with orthogonal polynomials and Riemann-Hilbert problems
• Singularity probability and the least singular value; connections with the Littlewood-Offord problem
• The circular law
• Universality for eigenvalue spacing; Erdos-Schlein-Yau delocalisation of eigenvectors and applications
If time permits, I may also cover
• The Tracy-Widom law
• Connections with Dyson Brownian motion and the Ornstein-Uhlenbeck process; the Erdos-Schlein-Yau approach to eigenvalue spacing universality
• Conjectural connections with zeroes of the Riemann zeta function
Depending on how the course progresses, I may also continue it into the spring quarter (or else have a spring graduate course on a different topic – one potential topic I have in mind is dynamics on nilmanifolds and applications to combinatorics).
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## my paper on the Born rule...
Hi,
A while ago I discussed here about a paper I wrote, which you can find on the arxiv: quant-ph/0505059
Proponents of the Everett interpretation of Quantum Theory have made efforts to show that to an observer in a branch, everything happens as if the projection postulate were true without postulating it. In this paper, we will indicate that it is only possible to deduce this rule if one introduces another postulate that is logically equivalent to introducing the projection postulate as an extra assumption. We do this by examining the consequences of changing the projection postulate into an alternative one, while keeping the unitary part of quantum theory, and indicate that this is a consistent (although strange) physical theory.
I submitted it to the Royal Society, and I received a notification of rejection, with the following comments from the referees, might be of interest for those who participated in the discussion. The emphasis is mine.
First referee:
The paper critically assesses the attempt (Proc Roy Soc Lond 1999) by David Deutsch (followed up by various authors in later work) to derive the Born rule within the Everett interpretation via considerations of decision theory. The author interprets Deutsch as claiming that QM - whether or not the Everett interpretation is assumed - may be decomposed into a unitary part and a "projection postulate" part. He then proposes an "alternative projection postulate" (APP) which, he argues, is compatible with the unitary part of QM but which does not entail the Born rule. He claims that, since his APP is a counterexample, Deutsch's proposal and any variants of it must be rejected. A very similar project was undertaken by Barnum et al in a paper in Proc Roy Soc Lond in 2000. The author's APP has some mild technical advantages over Barnum et al's proposal, but these do not (in my view) merit a separate paper, especially since neither Barnum et al nor the author are proposing a viable alternative to the PP but simply making a logical point. More importantly, the post-2000 literature on Deutsch's argument has not attempted to criticise the details of Barnum et al's counterexample. Rather, it has claimed that Barnum et al, treating measurement as a black-box process, misread Deutsch. Deutsch sets out to analyse measurement as one more physical process (realised within unitary dynamics - as such, any rival proposal to the Born rule which is couched (as is the author's) in terms of measurement observables taken as primitive will not be relevant within the context of the Everett interpretation. It is fair to say that this point was somewhat obscure in Deutsch's 1999 paper, but it has been made explicitly in subsequent discussions, including some (by Wallace and Greaves) which the author cites. However, the author does not engage with this issue but continues to work in the Barnum et al tradition without further discussion. In conclusion: if the Barnum et al framework is valid then the author's paper does not seem to add sufficiently to their existing criticisms of Deutsch to justify publication. And if it is not valid, then it is at best unclear how the author's paper relates to Deutsch.
The second referee:
The paper reviews an alternative projection postulate (APP) and contrasts it with the standard projection postulate (PP). Under the APP probabilities are uniform, instead of being proportional to the relative measure of vector components. APP is shown to be consistent with unitary symmetry and with measurements being defined in terms of projection operators, and it agrees with PP regarding results predicted with certainty. The paper also does a decent job of describing some of the strange empirical results that APP predicts. The main point, that we must rely on empirical data to favor PP over APP, is worth making. The paper, however, purports to do more than this. The abstract and introduction claim to deal a blow to the Everett programme, by showing that "there is no hope of deriving the PP directly from the rest of the machinery of quantum theory." Beyond the review of APP described above, however, the paper itself says very little about this subject. The introduction ends by promising "we will then examine where exactly it is in disagreement with Deutsch's reasonable assumptions,' or with Gleason's theorem." But the section at the end of the paper that is supposed to make good on this promise consists of only thirteen lines -- far too little to provide much exact examination. Worse, the paper does not mention or discuss any of the many other approaches that have been suggested for deriving the PP from the rest of quantum theory, within the Everett programme. The paper claims "APP is in fact the most natural probability rule that goes with the Everett interpretation: on each branching' of an observer due to a measurement, all of its alternative `worlds' receive and equal probability." However, many authors do not accept that equal probability per world is the most natural. Furthermore, many other authors do accept an equal probability rule, but then try to derive the PP from it, instead of the APP. For example, the review article at http://plato.stanford.edu/entries/qm-manyworlds/ says "Another idea for obtaining a probability law out of the formalism is to state, by analogy to the frequency interpretation of classical probability, that the probability of an outcome is proportional to the number of worlds with this outcome. This proposal immediately yields predictions that are different from what we observe in experiments. Some authors, arguing that counting is the only sensible way to introduce probability, consider this to be a fatal difficulty for the MWI, e.g., Belifante 1975. Graham 1973 suggested that the counting of worlds does yield correct probabilities if one takes into account detailed splitting of the worlds in realistic experiments, but other authors have criticized the MWI because of the failure of Graham's claim. Weissman 1999 has proposed a modification of quantum theory with additional non-linear decoherence (and hence even more worlds than standard MWI), which can lead asymptotically to worlds of equal mean measure for different outcomes." (Hanson 2003, which you incorrectly cite as discussing the Deutsch proof, is another such attempt.) I cannot recommend the paper for publication as it is, but I can hold out hope that the author could make an acceptable revision. Such a revision could simply be a review of the APP, including its implications. Such a review should mention many of the previous authors who have considered such a posulate. Alternatively, a revision could critique some of the attempts to derive PP from quantum theory. To accomplish this second goal, the author must first choose a set of previous papers that it is responding to. (It may not be feasible to respond to all previous papers on this topic.) Second, the author must explain exactly where there purported demonstration is claimed to fail. That is, at what point does a key assumption of theirs go beyond the basic machinery of quantum theory. Third, the author must explain why this key assumption is no more plausible than simply assuming the PP directly. This is what it would take to successfully show that such attempts to derive the PP from the machinery of quantum theory has failed. Here are two minor comments. The paper switches its notation from from APP to AQT and PP to SQT, for no apparent reason. It would make more sense to stick with one notation. Also, as there may be other alternatives proposed someday, it might be better to call APP a "uniform projection postulate" (UPP). Finally, the title should more specifically refer to this alternative projection posulate, however named.
On a personal note, although this paper was a bit outside of my field and thus "for fun", in my field too, I had several rejections of similar kind, which always make me think that the referee has missed the point I was trying to make (which must be due to the way I wrote it up, somehow).
The only point I tried to make was a logical one, as seems to be recognized by the first referee only, but then he seems to miss the point that in the end of the day, we want a theory that spits out results that are given by the PP, whether or not we take that "as primitive". So I don't see why considering the PP "as primitive" makes the reasoning "not relevant". The second referee seems to have understood this (that we have to rely on empirical data to endorse the PP), but he seems to have missed the point I was making a logical claim, and seems to concentrate on the minor remark when I said that "this APP seems to be the most natural probability rule going with MWI".
The very argument that some have tried to MODIFY QM introducing non-linear decoherence is *exactly what I claim*: that you need an extra hypothesis with unitary QM if you want to derive the PP. Finally, the proposition of revision, namely to limit myself to the consequences of the APP, take away the essential point of the paper which simply stated: since two different probability rules, the APP, and the PP, are both compatible with unitary QM, you cannot derive the PP logically from unitary QM without introducing an extra hypothesis.
The only truely valid critique I find here, is the one of the first referee who finds that my paper is not sufficiently different from Barnum's paper (something I ignored) - which is of course a valid reason of rejection (which I emphasised in red).
Most other points seem to miss the issue of the paper, I have the impression, and focus on details which are not relevant to the main point made. This often happens to me when I receive referee reports. Do others also have this impression, or am I such a terrible author ?
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All attempt to derive the PP from unitary theory is condemed to failure. It is a simple mathematical (and physical) question. The information contained into a nonunitary evolutor is more rich that informaiton contained into a unitary evolutor. 'More' cannot be derived from 'less'. It took near 100 years that physicists understood that measurement problem CANNOT be solved via QM of closed systems. During 50 years or so had an intensive research in open systems and decoherence. Finally decoherence is in a dead way. I wait that in some 100 or 200 years physicists will understand that the old unitary Schrödinger equation is an approximation to realistic nonunitary evolutions. In fact, in some other fields this is known for decades... See page 17 of Nobel Lecture, 8 December, 1977 http://nobelprize.org/chemistry/laur...ne-lecture.pdf The measurement process is an irreversible process generating entropy. QM conserves entropy and is reversible, therefore QM cannot explain the PP without invoking it adittionally. But then one is invoking a postulate that violates others postulates of the axiomatic structure, doing QM both incomplete and internally inconsistent.
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Quote by vanesch Hi, A while ago I discussed here about a paper I wrote, which you can find on the arxiv: quant-ph/0505059 I submitted it to the Royal Society, and I received a notification of rejection, with the following comments from the referees, might be of interest for those who participated in the discussion. The emphasis is mine. First referee: The second referee: On a personal note, although this paper was a bit outside of my field and thus "for fun", in my field too, I had several rejections of similar kind, which always make me think that the referee has missed the point I was trying to make (which must be due to the way I wrote it up, somehow). The only point I tried to make was a logical one, as seems to be recognized by the first referee only, but then he seems to miss the point that in the end of the day, we want a theory that spits out results that are given by the PP, whether or not we take that "as primitive". So I don't see why considering the PP "as primitive" makes the reasoning "not relevant". The second referee seems to have understood this (that we have to rely on empirical data to endorse the PP), but he seems to have missed the point I was making a logical claim, and seems to concentrate on the minor remark when I said that "this APP seems to be the most natural probability rule going with MWI". The very argument that some have tried to MODIFY QM introducing non-linear decoherence is *exactly what I claim*: that you need an extra hypothesis with unitary QM if you want to derive the PP. Finally, the proposition of revision, namely to limit myself to the consequences of the APP, take away the essential point of the paper which simply stated: since two different probability rules, the APP, and the PP, are both compatible with unitary QM, you cannot derive the PP logically from unitary QM without introducing an extra hypothesis. The only truely valid critique I find here, is the one of the first referee who finds that my paper is not sufficiently different from Barnum's paper (something I ignored) - which is of course a valid reason of rejection (which I emphasised in red). Most other points seem to miss the issue of the paper, I have the impression, and focus on details which are not relevant to the main point made. This often happens to me when I receive referee reports. Do others also have this impression, or am I such a terrible author ?
I always find that I have to ask an "outsider" to read my manuscript before I submit it. This is because what I find to be rather obvious, is really isn't. Authors have a clear idea in their heads what they're writing. Other people don't. So we tend to write things as if the reader already has an insight into our punch line. If you find that most people seem to miss the main point you're trying to make, chances are that you are not emphasizing it in the clearest fashion. This has happened even to the best of us.
I find that the most effective means to emphasize the main points I'm trying to get across is by clearly numbering them. I've been known on here to list the points one at a time:
(i) Point 1
(ii) Point 2
.. etc. Unfortunately, if you're writing to PRL, or trying to save publication costs, those take a lot of valuable spaces, so I also have listed them in line. As a referee, I also find them to be easier to focus on. I can easily go back and look at them again while I'm reading the rest of the paper to keep reminding myself that these are the points the authors are trying to make. It is no secret that most of us start a paper by reading the abstract, intro, and conclusion first (well, I certainly do). So you have to keep in mind that you literally have to reveal to the reader in the most direct way the message you are trying to get across in those sections of the paper.
Zz.
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## my paper on the Born rule...
Quote by ZapperZ I always find that I have to ask an "outsider" to read my manuscript before I submit it. This is because what I find to be rather obvious, is really isn't. Authors have a clear idea in their heads what they're writing. Other people don't. So we tend to write things as if the reader already has an insight into our punch line.
This reminds me of a simple paper I wrote once with a student, about how the signal generating process should be included in a reliable simulation of the behaviour of the front end electronics of a neutron detector, because assuming that the detector just "sent out a delta-pulse" was giving results which deviated by a factor of 2 from observations, while including the signal formation did predict this factor 2. I found this maybe worth publishing - even though not big news - because other papers omitted exactly that: they only took into account the electronics, and supposed a deltafunction for the signal coming from the detector (which might have been justified in their application, no matter - but it was not mentioned in their papers).
So I carefully described the setup, and gave an explicit calculation of how the signal was generated in the detector, to show that this was the relevant part which allowed us to explain the discrepancy of a factor of 2. My point being that it was necessary to include this part in the description.
I got a rather nasty referee report, in which he explained me that I must be pretty naive to think that I was the first one explaining how signals were generated in radiation detectors
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Quote by vanesch This reminds me of a simple paper I wrote once with a student, about how the signal generating process should be included in a reliable simulation of the behaviour of the front end electronics of a neutron detector, because assuming that the detector just "sent out a delta-pulse" was giving results which deviated by a factor of 2 from observations, while including the signal formation did predict this factor 2. I found this maybe worth publishing - even though not big news - because other papers omitted exactly that: they only took into account the electronics, and supposed a deltafunction for the signal coming from the detector (which might have been justified in their application, no matter - but it was not mentioned in their papers). So I carefully described the setup, and gave an explicit calculation of how the signal was generated in the detector, to show that this was the relevant part which allowed us to explain the discrepancy of a factor of 2. My point being that it was necessary to include this part in the description. I got a rather nasty referee report, in which he explained me that I must be pretty naive to think that I was the first one explaining how signals were generated in radiation detectors
I think it highly depends on WHERE you sent that in. If you sent it to, let's say, PRL, then I'd say you might get something like that. However, journals like EJP, or AJP, routinely publish pedagogy and techniques, especially when it is something relevant in physics education, be it at the undergraduate or graduate level.
I don't know what you submitted that paper to, but honestly, where you send your manuscript is almost as important as what you wrote.
Zz.
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Quote by ZapperZ I don't know what you submitted that paper to, but honestly, where you send your manuscript is almost as important as what you wrote. Zz.
It was Nuclear Instruments and Methods, quite appropriate, I'd think
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Quote by vanesch It was Nuclear Instruments and Methods, quite appropriate, I'd think
Well, I'm not sure about that.
NIM is supposed to be a journal on new techniques, or an improvement of a technique. Your paper, from your description, is simply clarifying the missing piece that isn't commonly mentioned. In other words, there's nothing new or a new extension on an existing technique. If this is the case, then the referee is correct in asking you if you think that what you're describing is not known.
I still think AJP or EJP might have been more suitable. You could emphasize the point that what you're describing is important and often omitted in the details of the experiment being reported in many papers that use the same technique. Such a paper would have been appropriate for those two journals.
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Quote by ZapperZ Well, I'm not sure about that. NIM is supposed to be a journal on new techniques, or an improvement of a technique. Your paper, from your description, is simply clarifying the missing piece that isn't commonly mentioned. In other words, there's nothing new or a new extension on an existing technique.
This is an interesting comment ! Nobody ever made that, and it explains several other problems I had with NIM ; indeed, each time I erred more on the explanatory part than the "here's a new method" part, I got rebiffed (or one asked me to remove or reduce the explanatory part and to emphasize the practical application). It is true that amongst my collegues, I'm by far the most "explanation" oriented.
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Quote by vanesch This is an interesting comment ! Nobody ever made that, and it explains several other problems I had with NIM ; indeed, each time I erred more on the explanatory part than the "here's a new method" part, I got rebiffed (or one asked me to remove or reduce the explanatory part and to emphasize the practical application). It is true that amongst my collegues, I'm by far the most "explanation" oriented.
I tend to be quite verbose too in some of the things I write. But as a referee, when I pick up a paper that I'm reviewing, I would like to be hit right off the bat with the punch line. Tell me in no uncertain terms what you are trying to say here, and why it is important. I tend to pay attention to statements such as these:
"To be best of our knowledge, this is the first report on.... "
"This results contradicts an earlier report...."
"This is the most accurate result so far on.... "
"This is a new result..... "
etc. These should be either in the abstract, or somewhere in the intro or the 1st 2 paragraph. If not, I will lose track of what you're trying to say, or why it is important. (Ironically, I've just finished reposting in my Journal an article I wrote a while back titled "It may be interesting, but is it important?") :)
If you write a paper in such a way that the referee has to put an effort to find the point you are making, or why it is important, then you are just making it more difficult for that referee to recommend your paper to be published. It is that simple.
Zz.
Quote by vanesch Hi, A while ago I discussed here about a paper I wrote, which you can find on the arxiv: quant-ph/0505059 I submitted it to the Royal Society, and I received a notification of rejection ...
Drat! I have my follow-up to your paper nearly ready for submission. Every weekend for the past several weeks now, I've told myself I'm going to make the final revisions and send it out, and then I run across something else that I need to incorporate. Like the Weissman paper, for instance ... In fact, I should probably make it at least evident that I'm aware of Weissman, Deutsch, Barnum, Hanson, and all the other authors mentioned in the reviews.
So Patrick, do you think you're going to resubmit? I hope you do - I think (obviously) that it is a very important topic. I'll try to throw out my comments on the reviewers' comments on this thread, as I go through the literature (may be a slow process ...)
BTW, does it normally take that long for review? Hasn't it been, what, 5 months?
David
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Quote by straycat ... In fact, I should probably make it at least evident that I'm aware of Weissman, Deutsch, Barnum, Hanson, and all the other authors mentioned in the reviews. So Patrick, do you think you're going to resubmit? I hope you do - I think (obviously) that it is a very important topic.
First I'll check out the Barnum paper. If (according to referee 1) my paper contains the same argument as his, well I conclude that 1) I'm in not a bad company (just 5 years late ) and 2) I won't resubmit.
If not, well, I wouldn't really know where to submit. Maybe foundations of physics.
BTW, does it normally take that long for review? Hasn't it been, what, 5 months? David
It's usually a bad sign when it takes that long But it is strongly dependent on the journal. Some journals first as one referee, and if that one doesn't give positive returns, they ask a second one for a second opinion. Others do it in parallel.
Quote by vanesch First I'll check out the Barnum paper. If (according to referee 1) my paper contains the same argument as his, well I conclude that 1) I'm in not a bad company (just 5 years late ) and 2) I won't resubmit.
A review article might not be such a bad idea. You could review the motivation behind the APP, review the various attempts to implement it, and perhaps include your own contribution in a separate section.
Quote by vanesch If not, well, I wouldn't really know where to submit. Maybe foundations of physics.
What exactly is the reputation of FoP? Is it a lesser tier than the Royal Society?
DS
Hey Patrick,
I've been trying to make sense of some of the comments made by your first referree:
Quote by vanesch A very similar project was undertaken by Barnum et al in a paper in Proc Roy Soc Lond in 2000. The author's APP has some mild technical advantages over Barnum et al's proposal, but these do not (in my view) merit a separate paper, especially since neither Barnum et al nor the author are proposing a viable alternative to the PP but simply making a logical point. More importantly, the post-2000 literature on Deutsch's argument has not attempted to criticise the details of Barnum et al's counterexample. Rather, it has claimed that Barnum et al, treating measurement as a black-box process, misread Deutsch. Deutsch sets out to analyse measurement as one more physical process (realised within unitary dynamics - as such, any rival proposal to the Born rule which is couched (as is the author's) in terms of measurement observables taken as primitive will not be relevant within the context of the Everett interpretation. It is fair to say that this point was somewhat obscure in Deutsch's 1999 paper, but it has been made explicitly in subsequent discussions, including some (by Wallace and Greaves) which the author cites.
I looked at one of Greaves' papers, "Understanding Deutsch's probability in a deterministic multiverse" which is archived at the PhilSci archives at http://philsci-archive.pitt.edu/archive/00001742/ . Section 5.1 "Measurement neutrality" and section 5.2: "Measurement Neutrality versus Egalitarianism" really helped me to understand the above point. Basically, Greaves explains that one of the essential assumptions in Deutsch-Wallace decision theory is the postulate of "measurement neutrality," which is "the assumption that a rational agent should be indifferent between any two quantum games that agree on the state |phi> to be measured, measurement operator X and payoff function P, regardless of how X is to me measured on |phi>." afaict, this means that if we think of the measurement process as a "black box," then Deutsch assumes that a rational agent should in principle be indifferent to the details of the innards of this black box.
In sec 5.2, Greaves very clearly argues that measurement neutrality automatically *excludes* the APP (where the APP = egalitarianism) as a possible probability rule. Therefore, measurement neutrality, as innocuous as it may appear at first glance, is not so innocuous at all.
I've referenced Greaves (among others) in the revised introduction to my paper [1] on the probability interpretation of the MWI. I'm glad you posted your referree comments, Patrick -- they've helped me on my paper!
-- David
[1] To be submitted to Foundations of Physics Letters -- latest draft available at
http://briefcase.yahoo.com/straycat_md
Folder: Probability interpretation of the MWI
archived (slightly older) versions also at:
http://philsci-archive.pitt.edu/
http://www.sciprint.org/
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Quote by straycat which is "the assumption that a rational agent should be indifferent between any two quantum games that agree on the state |phi> to be measured, measurement operator X and payoff function P, regardless of how X is to me measured on |phi>."
Yes, that's exactly the point. As I showed in my paper, that's NOT the case with the APP, (as I explicitly show with the example of X and Y where one is a refinement of the other) where the probabilities are dependent on context (on the other variables that are being measured).
In sec 5.2, Greaves very clearly argues that measurement neutrality automatically *excludes* the APP (where the APP = egalitarianism) as a possible probability rule. Therefore, measurement neutrality, as innocuous as it may appear at first glance, is not so innocuous at all.
Ok, that's exactly my argument too. So I have some extra homework to make with this as reference.
Thanks for pointing that out!
Juan wrote:
All attempt to derive the PP from unitary theory is condemed to failure.
I cannot agree with that statement, altough I recognize a conceptual difficulty there.
For me, this problem is similar to the problem of irreversibility seen from the classical mechanics point of view. Non-unitary evolution might be a good approximation (maybe even *exact*!) when an interaction with a huge system (huge freedom) is involved.
My favorite example is the decay of atomic states: clearly the interaction of the discrete atomic system with the continuum system of electromagnetic radiation brings the decay. This decay is very conveniently represented by a "non hermitian" hamiltonian: this allows modeling of an atom (for the Stark effect e.g.) without including the whole field. This represents correctly the reality, altough the fundamental laws are unitary.
For many people, the interaction with a 'classical' or 'macroscopic' system is all that is needed to derive the PP. I think this is the most probable explanation for the PP. Landau considered this so obvious that it comes in the first chapters in his QM book.
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Quote by lalbatros Juan wrote: I cannot agree with that statement, altough I recognize a conceptual difficulty there. For me, this problem is similar to the problem of irreversibility seen from the classical mechanics point of view.
The irreversibility in classical statistical mechanics comes about from the very specific initial condition, which is highly improbable.
Non-unitary evolution might be a good approximation (maybe even *exact*!) when an interaction with a huge system (huge freedom) is involved.
I don't see how this can come about. The hamiltonian gives rise to a unitary operator, no matter how complicated the system. Especially the EM radiation field can always be considered as a discrete system with a huge number of degrees of freedom (it shouldn't make any difference if you put your system in a box with diameter one hundred billion lightyears or not, should it).
My favorite example is the decay of atomic states: clearly the interaction of the discrete atomic system with the continuum system of electromagnetic radiation brings the decay. This decay is very conveniently represented by a "non hermitian" hamiltonian: this allows modeling of an atom (for the Stark effect e.g.) without including the whole field. This represents correctly the reality, altough the fundamental laws are unitary.
No, it is a shortcut, where you *apply* already the PP in its derivation.
For many people, the interaction with a 'classical' or 'macroscopic' system is all that is needed to derive the PP. I think this is the most probable explanation for the PP. Landau considered this so obvious that it comes in the first chapters in his QM book.
This is the standard "explanation". But it is *postulated* and not *derived* from unitary QM. What qualifies a system as "macroscopic" and "classical" (without making circular reasoning ?) and why shouldn't it obey standard quantum theory ?
Or is there an upper limit to the number product hilbert spaces (number of particles) before the exponentiation of a hermitean operator suddenly doesn't become unitary anymore ?
Hey Patrick et al, I'm posting an idea on this thread that has occurred to me on a potential consequence of the APP. Suppose that Alice is doing two Aspect-like experiments, one with Bob, and another simultaneously with Bert. Alice and Bob are 1 km apart, and Bert is 0.1 km farther away than Bob. Otherwise the experiments are the same, done at the same time. Bob and Bert flash the results of their measurements to Alice as soon as they get them. Before Alice receives these messages (which we suppose travel at the speed of light), Bob and Bert each exist in a superposition of the "B-- sees up"/"B-- sees down" state. Because of the general relativistic restriction on the speed of light, from the point of view of Alice, Bob's state will collapse prior to Bert's state. Pretty elementary. The point I wish to make is that relativity imposes a restriction on the order with which collapse occurs, from the point of view of Alice. So let's take this point and extrapolate. Suppose now that we have an observer Amandra who observes some variable X characteristic of a particle. But imagine that the value of X is not communicated to Amandra all at once, but rather in little chunks. That is, suppose that X_min is the lowest allowable value of X, and that it is quantized, ie it takes values in [X_min, X_min +1, X_min + 2, ...]. Imagine furthermore that Amandra's observation of X comes in a series of steps, like this: she observes either X = X_min, or X \in [X_min+1, X_min +2, ...]; if the latter, she next observes either X = X_min + 1 or X \in [X_min+2, X_min +3, ...]; if the latter, she next observes either X = X_min + 2, or X \in [X_min+3, X_min +4, ...]; and so on. If you draw the MWI-style world-splitting diagram to characterize this process **and apply the APP**, then it is apparent that lower possible values of X are *more probable* than higher values. In effect, X is MINIMIZED. We could equally well suppose that X might be maximized, if the information regarding the value of X were propagated to Amanda in the reverse order. So here's the Big Idea: the APP, perhaps, offers a mechanism by which Nature enforces various extremum laws. If X is the action, for instance, then we have the principle of least action. What d'ya think? David
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http://math.stackexchange.com/questions/238521/is-lim-n-to-infty-fracnn-frac1n-e-any-easier-than-stirling
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# Is $\lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e$ any easier than Stirling? [duplicate]
Possible Duplicate:
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$
Stirling's approximation says that $$\lim_{n \to \infty} \frac{n^n \sqrt{n}}{n! e^n } = \frac{1}{\sqrt{2 \pi}}.$$ Since $\lim_{n \to \infty} x^\frac{1}{n} \to 1$ uniformly on a neighbourhood of $\frac{1}{\sqrt{ 2 \pi}}$, it follows that $$\lim_{n \to \infty} \left( \frac{n^n \sqrt{n}}{n! e^n } \right)^\frac{1}{n} = \lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} \cdot \frac{n^\frac{1}{2n}}{e} = 1.$$ Since $\lim_{n \to \infty} n^\frac{1}{2n} = 1$, we get the limit in the title $$\lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e.$$
Question: Is Stirling's approximation is really needed to derive the above limit? Or is there an easier way to reach the same conclusion?
Motivation: The radius of convergence $R$ of a power series $\sum_{n=0}^\infty a_nx^n$ is given by Hadamard's formula $$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^\frac{1}{n}.$$ If we know ahead of time that $R > 0$ then the coefficients are given by $$a_n = \frac{f^{(n)}(0)}{n!}$$ where $f$ is the function defined by the power series. Then we get $$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^\frac{1}{n} = \limsup_{n \to \infty} \frac{|f^{(n)}(0)|^\frac{1}{n}}{n} \cdot \frac{n}{(n!)^\frac{1}{n}} = e \cdot \limsup_{n \to \infty} \frac{|f^{(n)}(0)|^\frac{1}{n}}{n}.$$ So one use of the limit is to clean up the formula for the radius of convergence of a power series in terms of the derivatives of the corresponding function.
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– sos440 Nov 16 '12 at 7:27
You only need Stirling's approximation up to a polynomial multiplicative constant, which you can get from a straightforward Riemann sum argument. – Qiaochu Yuan Nov 16 '12 at 8:20
## marked as duplicate by Did, TMM, Norbert, Arkamis, Chris EagleNov 16 '12 at 17:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 2 Answers
HINT: write $\lim_{n \to \infty} \displaystyle\frac{n}{(n!)^\frac{1}{n}}$ as $\lim_{n \to \infty} \left(\displaystyle\frac{n^n}{(n!)}\right)^\frac{1}{n}$ and then compute $\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}}$ where $a_n=\displaystyle\frac{n^n}{(n!)}$
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Taking the natural logarithm, you want to show that $$\lim_{n\to\infty}\left(\ln n-\frac1{n}\sum_{i=1}^n\ln i\right)=1$$ or equivalently $$n \ln n - \sum_{i=1}^n\ln i \sim n\qquad\text{as }n\to\infty.$$ Now $$\int_1^n\ln x\,dx<\sum_{i=1}^n\ln i<\int_1^n\ln x\,dx+\ln n,$$ and since $x\ln x-x$ is a primitive of $\ln x$, it follows that $$\sum_{i=1}^n\ln i = n \ln n-n+O(\ln n)$$ which approximation suffices easily for the asymptotic equivalence above.
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an alternative from the $2$nd line is: $\lim_{n\to\infty}\left(\ln n-\frac1{n}\sum_{i=1}^n\ln i\right)=\lim_{n\to\infty}\left(\ln n-\left(\frac1{n}\sum_{i=1}^n\ln \frac{i}{n}+\ln n\right)\right)=-\int_0^1 \ln x=1.$ – Chris's wise sister Nov 16 '12 at 11:53
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http://en.wikipedia.org/wiki/T-symmetry
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# T-symmetry
In theoretical physics, T-symmetry is the theoretical symmetry of physical laws under a time reversal transformation:
$T: t \mapsto -t.$
Although in restricted contexts one may find this symmetry, the observable universe itself does not show symmetry under time reversal, primarily due to the second law of thermodynamics. Hence time is said to be non-symmetric, or asymmetric.
Time asymmetries are generally distinguished as between those intrinsic to the dynamic laws of nature, and those due to the initial conditions of our universe.
1. The T-asymmetry of the weak force is of the first kind, while
2. the T-asymmetry of the second law of thermodynamics is of the second kind.
## Invariance
Physicists also discuss the time-reversal invariance of local and/or macroscopic descriptions of physical systems, independent of the invariance of the underlying microscopic physical laws. For example, Maxwell's equations with material absorption or Newtonian mechanics with friction are not time-reversal invariant at the macroscopic level where they are normally applied, even if they are invariant at the microscopic level when one includes the atomic motions the "lost" energy is translated into.
A toy called the teeter-totter illustrates the two aspects of time reversal invariance. When set into motion atop a pedestal, the figure oscillates for a very long time. The toy is engineered to minimize friction and illustrate the reversibility of Newton's laws of motion. However, the mechanically stable state of the toy is when the figure falls down from the pedestal into one of arbitrarily many positions. This is an illustration of the law of increase of entropy through Boltzmann's identification of the logarithm of the number of states with the entropy.
## Macroscopic phenomena: the second law of thermodynamics
Our daily experience shows that T-symmetry does not hold for the behavior of bulk materials. Of these macroscopic laws, most notable is the second law of thermodynamics. Many other phenomena, such as the relative motion of bodies with friction, or viscous motion of fluids, reduce to this, because the underlying mechanism is the dissipation of usable energy (for example, kinetic energy) into heat.
The question of whether this time-asymmetric dissipation is really inevitable has been considered by many physicists, often in the context of Maxwell's demon. The name comes from a thought experiment described by James Clerk Maxwell in which a microscopic demon guards a gate between two halves of a room. It only lets slow molecules into one half, only fast ones into the other. By eventually making one side of the room cooler than before and the other hotter, it seems to reduce the entropy of the room, and reverse the arrow of time. Many analyses have been made of this; all show that when the entropy of room and demon are taken together, this total entropy does increase. Modern analyses of this problem have taken into account Claude E. Shannon's relation between entropy and information. Many interesting results in modern computing are closely related to this problem — reversible computing, quantum computing and physical limits to computing, are examples. These seemingly metaphysical questions are today, in these ways, slowly being converted to the stuff of the physical sciences.
The current consensus hinges upon the Boltzmann-Shannon identification of the logarithm of phase space volume with the negative of Shannon information, and hence to entropy. In this notion, a fixed initial state of a macroscopic system corresponds to relatively low entropy because the coordinates of the molecules of the body are constrained. As the system evolves in the presence of dissipation, the molecular coordinates can move into larger volumes of phase space, becoming more uncertain, and thus leading to increase in entropy.
One can, however equally well imagine a state of the universe in which the motions of all of the particles at one instant were the reverse (strictly, the CPT reverse). Such a state would then evolve in reverse, so presumably entropy would decrease (Loschmidt's paradox). Why is 'our' state preferred over the other?
One position is to say that the constant increase of entropy we observe happens only because of the initial state of our universe. Other possible states of the universe (for example, a universe at heat death equilibrium) would actually result in no increase of entropy. In this view, the apparent T-asymmetry of our universe is a problem in cosmology: why did the universe start with a low entropy? This view, if it remains viable in the light of future cosmological observation, would connect this problem to one of the big open questions beyond the reach of today's physics — the question of initial conditions of the universe.
## Macroscopic phenomena: black holes
An object can cross through the event horizon of a black hole from the outside, and then fall rapidly to the central region where our understanding of physics breaks down. Since within a black hole the forward light-cone is directed towards the center and the backward light-cone is directed outward, it is not even possible to define time-reversal in the usual manner. The only way anything can escape from a black hole is as Hawking radiation.
The time reversal of a black hole would be a hypothetical object known as a white hole. From the outside they appear similar. While a black hole has a beginning and is inescapable, a white hole has an ending and cannot be entered. The forward light-cones of a white hole are directed outward; and its backward light-cones are directed towards the center.
The event horizon of a black hole may be thought of as a surface moving outward at the local speed of light and is just on the edge between escaping and falling back. The event horizon of a white hole is a surface moving inward at the local speed of light and is just on the edge between being swept outward and succeeding in reaching the center. They are two different kinds of horizons—the horizon of a white hole is like the horizon of a black hole turned inside-out.
The modern view of black hole irreversibility is to relate it to the second law of thermodynamics, since black holes are viewed as thermodynamic objects. Indeed, according to the Gauge-gravity duality conjecture, all microscopic processes in a black hole are reversible, and only the collective behavior is irreversible, as in any other macroscopic, thermal system.[citation needed]
## Kinetic consequences: detailed balance and Onsager reciprocal relations
In physical and chemical kinetics, T-symmetry of the mechanical microscopic equations implies two important laws: the principle of detailed balance and the Onsager reciprocal relations. T-symmetry of the microscopic description together with its kinetic consequences are called microscopic reversibility.
## Effect of time reversal on some variables of classical physics
### Even
Classical variables that do not change upon time reversal include:
$\vec x\!$, Position of a particle in three-space
$\vec a\!$, Acceleration of the particle
$\vec F\!$, Force on the particle
$E\!$, Energy of the particle
$\phi\!$, Electric potential (voltage)
$\vec E\!$, Electric field
$\vec D\!$, Electric displacement
$\rho\!$, Density of electric charge
$\vec P\!$, Electric polarization
Energy density of the electromagnetic field
Maxwell stress tensor
All masses, charges, coupling constants, and other physical constants, except those associated with the weak force.
### Odd
Classical variables that time reversal negates include:
$t\!$, The time when an event occurs
$\vec v\!$, Velocity of a particle
$\vec p\!$, Linear momentum of a particle
$\vec l\!$, Angular momentum of a particle (both orbital and spin)
$\vec A\!$, Electromagnetic vector potential
$\vec B\!$, Magnetic induction
$\vec H\!$, Magnetic field
$\vec j\!$, Density of electric current
$\vec M\!$, Magnetization
$\vec S\!$, Poynting vector
Power (rate of work done).
## Microscopic phenomena: time reversal invariance
Since most systems are asymmetric under time reversal, it is interesting to ask whether there are phenomena that do have this symmetry. In classical mechanics, a velocity v reverses under the operation of T, but an acceleration does not. Therefore, one models dissipative phenomena through terms that are odd in v. However, delicate experiments in which known sources of dissipation are removed reveal that the laws of mechanics are time reversal invariant. Dissipation itself is originated in the second law of thermodynamics.
The motion of a charged body in a magnetic field, B involves the velocity through the Lorentz force term v×B, and might seem at first to be asymmetric under T. A closer look assures us that B also changes sign under time reversal. This happens because a magnetic field is produced by an electric current, J, which reverses sign under T. Thus, the motion of classical charged particles in electromagnetic fields is also time reversal invariant. (Despite this, it is still useful to consider the time-reversal non-invariance in a local sense when the external field is held fixed, as when the magneto-optic effect is analyzed. This allows one to analyze the conditions under which optical phenomena that locally break time-reversal, such as Faraday isolators, can occur.) The laws of gravity also seem to be time reversal invariant in classical mechanics.
In physics one separates the laws of motion, called kinematics, from the laws of force, called dynamics. Following the classical kinematics of Newton's laws of motion, the kinematics of quantum mechanics is built in such a way that it presupposes nothing about the time reversal symmetry of the dynamics. In other words, if the dynamics are invariant, then the kinematics will allow it to remain invariant; if the dynamics is not, then the kinematics will also show this. The structure of the quantum laws of motion are richer, and we examine these next.
### Time reversal in quantum mechanics
Two-dimensional representations of parity are given by a pair of quantum states that go into each other under parity. However, this representation can always be reduced to linear combinations of states, each of which is either even or odd under parity. One says that all irreducible representations of parity are one-dimensional. Kramers' theorem states that time reversal need not have this property because it is represented by an anti-unitary operator.
This section contains a discussion of the three most important properties of time reversal in quantum mechanics; chiefly,
1. that it must be represented as an anti-unitary operator,
2. that it protects non-degenerate quantum states from having an electric dipole moment,
3. that it has two-dimensional representations with the property T2 = −1.
The strangeness of this result is clear if one compares it with parity. If parity transforms a pair of quantum states into each other, then the sum and difference of these two basis states are states of good parity. Time reversal does not behave like this. It seems to violate the theorem that all abelian groups be represented by one dimensional irreducible representations. The reason it does this is that it is represented by an anti-unitary operator. It thus opens the way to spinors in quantum mechanics.
### Anti-unitary representation of time reversal
Eugene Wigner showed that a symmetry operation S of a Hamiltonian is represented, in quantum mechanics either by a unitary operator, S = U, or an antiunitary one, S = UK where U is unitary, and K denotes complex conjugation. These are the only operations that acts on Hilbert space so as to preserve the length of the projection of any one state-vector onto another state-vector.
Consider the parity operator. Acting on the position, it reverses the directions of space, so that P−1xP = −x. Similarly, it reverses the direction of momentum, so that PpP−1 = −p, where x and p are the position and momentum operators. This preserves the canonical commutator [x, p] = iħ, where ħ is the reduced Planck constant, only if P is chosen to be unitary, PiP−1 = i.
On the other hand, for time reversal, the time-component of the momentum is the energy. If time reversal were implemented as a unitary operator, it would reverse the sign of the energy just as space-reversal reverses the sign of the momentum. This is not possible, because, unlike momentum, energy is always positive. Since energy in quantum mechanics is defined as the phase factor exp(-iEt) that one gets when one moves forward in time, the way to reverse time while preserving the sign of the energy is to reverse the sense of "i", so that the sense of phases is reversed.
Similarly, any operation that reverses the sense of phase, which changes the sign of i, will turn positive energies into negative energies unless it also changes the direction of time. So every antiunitary symmetry in a theory with positive energy must reverse the direction of time. The only antiunitary symmetry is time reversal, together with a unitary symmetry that does not reverse time.
Given the time reversal operator T, it does nothing to the x-operator, TxT−1 = x, but it reverses the direction of p, so that TpT−1 = −p. The canonical commutator is invariant only if T is chosen to be anti-unitary, i.e., TiT−1 = −i. For a particle with spin J, one can use the representation
$T = e^{-i\pi J_y/\hbar} K,$
where Jy is the y-component of the spin, and use of TJT−1 = −J has been made.
### Electric dipole moments
This has an interesting consequence on the electric dipole moment (EDM) of any particle. The EDM is defined through the shift in the energy of a state when it is put in an external electric field: Δe = d·E + E·δ·E, where d is called the EDM and δ, the induced dipole moment. One important property of an EDM is that the energy shift due to it changes sign under a parity transformation. However, since d is a vector, its expectation value in a state |ψ> must be proportional to <ψ| J |ψ>. Thus, under time reversal, an invariant state must have vanishing EDM. In other words, a non-vanishing EDM signals both P and T symmetry-breaking.
It is interesting to examine this argument further, since one feels that some molecules, such as water, must have EDM irrespective of whether T is a symmetry. This is correct: if a quantum system has degenerate ground states that transform into each other under parity, then time reversal need not be broken to give EDM.
Experimentally observed bounds on the electric dipole moment of the nucleon currently set stringent limits on the violation of time reversal symmetry in the strong interactions, and their modern theory: quantum chromodynamics. Then, using the CPT invariance of a relativistic quantum field theory, this puts strong bounds on strong CP violation.
Experimental bounds on the electron electric dipole moment also place limits on theories of particle physics and their parameters.
### Kramers' theorem
Main article: Kramers' degeneracy theorem
For T, which is an anti-unitary Z2 symmetry generator
T2 = UKUK = U U* = U (UT)−1 = Φ,
where Φ is a diagonal matrix of phases. As a result, U = ΦUT and UT = UΦ, showing that
U = Φ U Φ.
This means that the entries in Φ are ±1, as a result of which one may have either T2 = ±1. This is specific to the anti-unitarity of T. For a unitary operator, such as the parity, any phase is allowed.
Next, take a Hamiltonian invariant under T. Let |a> and T|a> be two quantum states of the same energy. Now, if T2 = −1, then one finds that the states are orthogonal: a result called Kramers' theorem. This implies that if T2 = −1, then there is a twofold degeneracy in the state. This result in non-relativistic quantum mechanics presages the spin statistics theorem of quantum field theory.
Quantum states that give unitary representations of time reversal, i.e., have T2=1, are characterized by a multiplicative quantum number, sometimes called the T-parity.
Time reversal transformation for fermions in quantum field theories can be represented by an 8-component spinor in which the above mentioned T-parity can be a complex number with unit radius. The CPT invariance is not a theorem but a better to have property in these class of theories.
### Time reversal of the known dynamical laws
Particle physics codified the basic laws of dynamics into the standard model. This is formulated as a quantum field theory that has CPT symmetry, i.e., the laws are invariant under simultaneous operation of time reversal, parity and charge conjugation. However, time reversal itself is seen not to be a symmetry (this is usually called CP violation). There are two possible origins of this asymmetry, one through the mixing of different flavours of quarks in their weak decays, the second through a direct CP violation in strong interactions. The first is seen in experiments, the second is strongly constrained by the non-observation of the EDM of a neutron.
It is important to stress that this time reversal violation is unrelated to the second law of thermodynamics, because due to the conservation of the CPT symmetry, the effect of time reversal is to rename particles as antiparticles and vice versa. Thus the second law of thermodynamics is thought to originate in the initial conditions in the universe.
## See also
• The second law of thermodynamics, Maxwell's demon and the arrow of time (also Loschmidt's paradox).
• Microscopic reversibility
• Detailed balance
• Applications to reversible computing and quantum computing, including limits to computing.
• The standard model of particle physics, CP violation, the CKM matrix and the strong CP problem
• Neutrino masses and CPT invariance.
• Wheeler–Feynman absorber theory
• Teleonomy
## References
• Maxwell's demon: entropy, information, computing, edited by H.S.Leff and A.F. Rex (IOP publishing, 1990) [ISBN 0-7503-0057-4]
• Maxwell's demon, 2: entropy, classical and quantum information, edited by H.S.Leff and A.F. Rex (IOP publishing, 2003) [ISBN 0-7503-0759-5]
• The emperor's new mind: concerning computers, minds, and the laws of physics, by Roger Penrose (Oxford university press, 2002) [ISBN 0-19-286198-0]
• Sozzi, M.S. (2008). Discrete symmetries and CP violation. Oxford University Press. ISBN 978-0-19-929666-8.
• Birss, R. R. (1964). Symmetry and Magnetism. John Wiley & Sons, Inc., New York.
• CP violation, by I.I. Bigi and A.I. Sanda (Cambridge University Press, 2000) [ISBN 0-521-44349-0]
• Particle Data Group on CP violation
• the Babar experiment in SLAC
• the BELLE experiment in KEK
• the KTeV experiment in Fermilab
• the CPLEAR experiment in CERN
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http://math.stackexchange.com/questions/242537/calculate-asymptotes-and-local-extreme-values
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# Calculate asymptotes and local extreme values
I'm fed up with this question from my book. I've calculated the constants to this equation but got stuck at the asymptotes and local extreme values calculations which I need to plot the graph, perhaps anyone could help me out or guide me towards the solution of calculating the asymptotes/local extreme values and then to plot the graph.
Equation:
Define the constants A,B,C so that a function which is defined by
```` (1) (6/pi) arctan(2-(x+2)²) when x < -1
f(x) = (2) x + c* |x| - 1 when -1 ≥ x ≥ 1
(3) (1/Ax+B) + 4 when x > 1 och Ax + B ≠ 0
````
is continuous at x = -1 and differentiable in x = 1
I calculated the constants, A,B,C to:
A = -18
B = 16
C = 7/2
Any help is appreciated,
Thanks, Michael.
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– Berci Nov 22 '12 at 12:59
## 1 Answer
Your values for $A,B,C$ are right, at least I got the same values to match the continuity and derivative conditions.
The top arctan function has a single critical point at $x=-2$. This is because the derivative of $\arctan u$ is $1/(1+u^2) \cdot u'$, so the only critical points come from the derivative of the "inside" function, in your case $2-(x+2)^2=-x^2-4x-2.$ This critical point is a local maximum of the top function.
The middle function doesn't have critical points at which derivative is zero, but the value $x=0$ is to be considered a critical point since the derivative of $|x|$ is undefined there. So the graph on the middle part is V shaped, local min at 0.
Neither the top nor the middle have vertical asymptotes, but the bottom function, which by the way has no critical points, has a vertical asymptote where $Ax+B=0$, i.e. at $x=8/9$.
Finally for horizontal asymptotes you can use that arctan itself has asymptote $-\pi/2$ going toward $- \infty$, but the function is multiplied by $6/\pi$ so combine these facts.
And the bottom function, which applies for large positive $x$, has asymptote $y=4$.
EDIT: You should also check for critical points at $x=-1,1$ since it's a piecewise function with functions on pieces changing at these points.
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Note it should also be checked at -1 and 1. I inserted that in the above answer... – coffeemath Nov 22 '12 at 14:53
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http://www.physicsforums.com/showthread.php?p=3768147
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Physics Forums
## Change in Density due to Temperature and Pressure Change
1. The problem statement, all variables and given/known data
Estimate the percent change in density of iron when it is still a solid, but deep in the Earth where the temperature is $2000^\circ C$ and it is under $5000$ atm of pressure. Take into account both thermal expansion and changes due to increased outside pressure. Assume both the bulk modulus and the volume coefficient of expansion do not vary with temperature and are the same as at normal room temperature. The bulk modulus for iron is about $90\times10^9N/m^2$.
We also know that the volumetric expansion coefficient of iron is $35\times10^{-6}(^{\circ}C)^{-1}$.
Also, $1atm=1.103\times10^5N/m^2$.
2. Relevant equations
$\frac{\Delta V}{V_0}=-\frac{1}{B}\Delta P$ where B is the bulk modulus.
$\Delta V=V_0\beta\Delta T$ where $\beta$ is the volumetric expansion coefficient.
3. The attempt at a solution
I'm assuming the change in volumes adds up so we have using the two equations successively, we have
$\frac{\Delta V}{V_0}=\left(-\frac{1}{90\times10^9}\right)\left(5000atm\right) \left(\frac{1.103\times10^5N/m^2}{1atm}\right)=-0.00563$
and
$\frac{\Delta V}{V_0}=(4975^\circ C)(35\times10^-6\left(^\circ C\right)^{-1}=0.17413$
So the total change in volume is $0.17413-0.00563=0.168495$.
Then we could easily get the percent change in density. However, I am unsure if I am applying this correctly. Am I allowed to add the change in volumes?
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions:
Homework Help
Quote by JSGandora However, I am unsure if I am applying this correctly. Am I allowed to add the change in volumes?
The volume becomes (1+ΔV/V0) times the initial one in both processes, and you need to multiply them when there is change of volume both because of pressure and temperature. When ΔV/V0<<1, you can simply add the relative changes (with the proper signs) but 0.17 is not very small compared to unity.
ehild
Wait, so it should be (0.17413)(-0.00563)=-0.0009803519? That doesn't make intuitive sense to me. Why do we multiply the change in volumes?
Recognitions:
Homework Help
## Change in Density due to Temperature and Pressure Change
Quote by JSGandora Wait, so it should be (0.17413)(-0.00563)=-0.0009803519?
NO. We multiply the volume ratios. V/V0=1.17413*(1-0.00563)=1.16752. ΔV/V0=0.16752.
ehild
Oh, I see now. Could you explain why it must be multiplied? I don't understand.
Recognitions: Homework Help The formulae for volume change due to pressure and temperature refer to cases when the other quantity is constant. Pretend you put the piece of iron at high pressure first, when the volume decreases by Vo(ΔP/B). The new volume, V1=Vo(1-ΔP/B) is warmed up, and the new volume serves as "Vo" in thermal expansion. But the results are almost the same, and the the formulae are approximations anyway. So both ways should be accepted for the relative change of the volume, which is 0.168 with 3 significant digits by both methods. ehild
Oh, thanks you so much. I understand now.
Tags
expansion, heat, pressure, temperature, thermodynamics
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http://mathoverflow.net/questions/86058/saturated-ideals
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## saturated ideals
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is it possible to have a saturated ideal on a successor cardinal which does not extend the nonstationary ideal? (i.e. some nonstationary set is positive for this ideal)
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## 1 Answer
Yes, it is. The reason is that an ideal $I$ on $P(\kappa)$ is saturated just in case the quotient Boolean algebra $P(\kappa)/I$ satisfies the $\kappa^+$-chain condition, and this is a property that is preserved by permutations of the underlying set $\kappa$. But the property of extending the non-stationary ideal is not preserved by such permutations, since we can perform a permutation of $\kappa$ that takes a nonstationary set to a club set. Thus, there are isomorphic versions of any saturated ideal that remain saturated, but which do not extend the non-stationary ideal.
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But if you add a normality condition, the story becomes more complicated. – Joel David Hamkins Jan 19 2012 at 1:42
I also just realized, if you take the induced ideal: $\{ X : 1 \Vdash \delta \in j(X) \}$, then this is saturated for any choice of delta, so we may pick a successor ordinal less than $\kappa^+$ and the get set of limit ordinals to be in the induced ideal. – mbsq Jan 19 2012 at 1:44
Yes, I think that is right. – Joel David Hamkins Jan 19 2012 at 1:48
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http://math.stackexchange.com/questions/11026/inverse-of-a-monoid-homomorphism
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# Inverse of a monoid homomorphism
It is a well-known fact that
$f\colon a \mapsto \left(\begin{array}{cc} 1 & 1\\ 0 & 1\\ \end{array}\right), ~ b \mapsto \left(\begin{array}{cc} 1 & 0\\ 1 & 1\\ \end{array}\right)$
is a monomorphism from the free monoid generated by $a$ and $b$ to the matrix monoid $\mathbb{Z}^{2\times 2}$.
1. Is there an efficient algorithm which computes the length of $f^{-1}(X)$ from $X \in \mathbb{Z}^{2\times 2}$ (that is, the number of symbols in corresponding word $x$, if $f^{-1}(X) = \{x\}$)?
2. Is there an efficient algorithm which computes $f^{-1}(X)$ from $X \in \mathbb{Z}^{2\times 2}$?
3. Is there a standard reference for this problem?
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## 1 Answer
Yes; see the Wikipedia article on Smith normal form. In two dimensions you are essentially using the Euclidean algorithm.
Edit: The basic observation here is that if $M = \left[ \begin{array}{cc} x & y \\\ z & w \end{array} \right]$ is an integer matrix, then
$$M f(a^{-1}) = \left[ \begin{array}{cc} x & y - z \\\ z & w - z \end{array} \right]$$
and similarly
$$M f(b^{-1}) = \left[ \begin{array}{cc} x - y & y \\\ z - w & w \end{array} \right].$$
If in addition $\det M = 1$ and $x, y, z, w$ are non-negative integers, then exactly one of these operations will give a matrix with the same properties (this is equivalent to $f$ being a monomorphism). So, as I said, essentially one is performing the Euclidean algorithm on the columns (or, if you multiply from the left instead, on the rows).
As an example, if $M = \left[ \begin{array}{cc} 8 & 5 \\\ 3 & 2 \end{array} \right]$, then
$$M f(b^{-1}) = \left[ \begin{array}{cc} 3 & 5 \\\ 1 & 2 \end{array} \right]$$
$$M f(b^{-1} a^{-1}) = \left[ \begin{array}{cc} 3 & 2 \\\ 1 & 1 \end{array} \right]$$
$$M f(b^{-1} a^{-1} b^{-1}) = \left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right] = f(aa)$$
hence $M = f(aabab)$ as desired.
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I recall it from a proof of structure theorem for finitely generated modules over PIDs. But I don't see how to apply it here. Could you elaborate on this? – Alexis Nov 20 '10 at 0:53
The algorithm for writing a matrix in Smith normal form is precisely the algorithm you want. Like I said, it's essentially the Euclidean algorithm. You can also think about it as a form of row reduction. – Qiaochu Yuan Nov 20 '10 at 11:20
f (aabab) = [[8,5],[3,2]]. Its SNF is [[1,0],[0,1]]. How does it help to recover aabab (or its length = 5) from [[8,5],[3,2]]? – Alexis Nov 20 '10 at 12:41
@Alexis: you need to keep track of the intermediate steps, not just the final answer. I will edit with an example. – Qiaochu Yuan Nov 20 '10 at 13:33
Thanks! – Alexis Nov 20 '10 at 13:43
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Probability
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# All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary
# Science Fair Project Encyclopedia
For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.
# Probability
The word probability derives from the Latin probare (to prove, or to test). Informally, probable is one of several words applied to uncertain events or knowledge, being more or less interchangeable with likely, risky, hazardous, uncertain, and doubtful, depending on the context. Chance, odds, and bet are other words expressing similar notions. As with the theory of mechanics which assigns precise definitions to such everyday terms as work and force, so the theory of probability attempts to quantify the notion of probable.
Contents
## Historical remarks
The scientific study of probability is a modern development. Gambling shows that there has been an interest in quantifying the ideas of probability for millennia, but exact mathematical descriptions of use in those problems only arose much later.
The doctrine of probabilities dates to the correspondence of Pierre de Fermat and Blaise Pascal (1654). Christiaan Huygens (1657) gave the first scientific treatment of the subject. Jakob Bernoulli's Ars Conjectandi (posthumous, 1713) and Abraham de Moivre's Doctrine of Chances (1718) treated the subject as a branch of mathematics.
The theory of errors may be traced back to Roger Cotes's Opera Miscellanea (posthumous, 1722), but a memoir prepared by Thomas Simpson in 1755 (printed 1756) first applied the theory to the discussion of errors of observation. The reprint (1757) of this memoir lays down the axioms that positive and negative errors are equally probable, and that there are certain assignable limits within which all errors may be supposed to fall; continuous errors are discussed and a probability curve is given.
Pierre-Simon Laplace (1774) made the first attempt to deduce a rule for the combination of observations from the principles of the theory of probabilities. He represented the law of probability of errors by a curve y = φ(x), x being any error and y its probability, and laid down three properties of this curve: (1) It is symmetric as to the y-axis; (2) the x-axis is an asymptote, the probability of the error $\infty$ being 0; (3) the area enclosed is 1, it being certain that an error exists. He deduced a formula for the mean of three observations. He also gave (1781) a formula for the law of facility of error (a term due to Lagrange, 1774), but one which led to unmanageable equations. Daniel Bernoulli (1778) introduced the principle of the maximum product of the probabilities of a system of concurrent errors.
The method of least squares is due to Adrien-Marie Legendre (1805), who introduced it in his Nouvelles méthodes pour la détermination des orbites des comètes. In ignorance of Legendre's contribution, an Irish-American writer, Robert Adrain, editor of "The Analyst" (1808), first deduced the law of facility of error,
$\phi(x) = ce^{-h^2 x^2}$
c and h being constants depending on precision of observation. He gave two proofs, the second being essentially the same as John Herschel's (1850). Gauss gave the first proof which seems to have been known in Europe (the third after Adrain's) in 1809. Further proofs were given by Laplace (1810, 1812), Gauss (1823), James Ivory (1825, 1826), Hagen (1837), Friedrich Bessel (1838), Donkin (1844, 1856), and Morgan Crofton (1870). Other contributors were Ellis (1844), De Morgan (1864), Glaisher (1872), and Giovanni Schiaparelli (1875). Peters's (1856) formula for r, the probable error of a single observation, is well known.
In the nineteenth century authors on the general theory included Laplace, Sylvestre Lacroix (1816), Littrow (1833), Adolphe Quetelet (1853), Richard Dedekind (1860), Helmert (1872), Hermann Laurent (1873), Liagre, Didion, and Karl Pearson. Augustus De Morgan and George Boole improved the exposition of the theory.
On the geometric side (see integral geometry) contributors to The Educational Times were influential (Miller, Crofton, McColl, Wolstenholme, Watson, and Artemas Martin).
## Concepts
There is essentially one set of mathematical rules for manipulating probability; these rules are listed under "Formalization of probability" below. (There are other rules for quantifying uncertainty, such as the Dempster-Shafer theory and fuzzy logic, but those are essentially different and not compatible with the laws of probability as they are usually understood.) However, there is ongoing debate over what, exactly, the rules apply to; this is the topic of probability interpretations.
The general idea of probability is often divided into two related concepts:
• Aleatory probability, which represents the likelihood of future events whose occurrence is governed by some random physical phenomenon. This concept can be further divided into physical phenomena that are predictable, in principle, with sufficient information, and phenomena which are essentially unpredictable. Examples of the first kind include tossing dice or spinning a roulette wheel, and an example of the second kind is radioactive decay.
• Epistemic probability, which represents our uncertainty about propositions when one lacks complete knowledge of causative circumstances. Such propositions may be about past or future events, but need not be. Some examples of epistemic probability are to assign a probability to the proposition that a proposed law of physics is true, and to determine how "probable" it is that a suspect committed a crime, based on the evidence presented.
It is an open question whether aleatory probability is reducible to epistemic probability based on our inability to precisely predict every force that might affect the roll of a die, or whether such uncertainties exist in the nature of reality itself, particularly in quantum phenomena governed by Heisenberg's uncertainty principle. Although the same mathematical rules apply regardless of which interpretation is chosen, the choice has major implications for the way in which probability is used to model the real world.
## Formalization of probability
Like other theories, the theory of probability is a representation of probabilistic concepts in formal terms -- that is, in terms that can be considered separately from their meaning. These formal terms are manipulated by the rules of mathematics and logic, and any results are then interpreted or translated back into the problem domain.
There have been at least two successful attempts to formalize probability, namely the Kolmogorov formulation and the Cox formulation. In Kolmogorov's formulation, sets are interpreted as events and probability itself as a measure on a class of sets. In Cox's formulation, probability is taken as a primitive (that is, not further analyzed) and the emphasis is on constructing a consistent assignment of probability values to propositions. In both cases, the laws of probability are the same, except for technical details:
1. a probability is a number between 0 and 1;
2. the probability of an event or proposition and its complement must add up to 1; and
3. the joint probability of two events or propositions is the product of the probability of one of them and the probability of the second, conditional on the first.
The reader will find an exposition of the Kolmogorov formulation in the probability theory article, and in the Cox's theorem article for Cox's formulation. See also the article on probability axioms.
For an algebraic alternative to Kolmogorov's approach, see algebra of random variables.
### Representation and interpretation of probability values
The probability of an event is generally represented as a real number between 0 and 1, inclusive. An impossible event has a probability of exactly 0, and a certain event has a probability of 1, but the converses are not always true: probability 0 events are not always impossible, nor probability 1 events certain. The rather subtle distinction between "certain" and "probability 1" is treated at greater length in the article on "almost surely".
Most probabilities that occur in practice are numbers between 0 and 1, indicating the event's position on the continuum between impossibility and certainty. The closer an event's probability is to 1, the more likely it is to occur.
For example, if two mutually exclusive events are assumed equally probable, such as a flipped coin landing heads-up or tails-up, we can express the probability of each event as "1 in 2", or, equivalently, "50%" or "1/2".
Probabilities are equivalently expressed as odds, which is the ratio of the probability of one event to the probability of all other events. The odds of heads-up, for the tossed coin, are (1/2)/(1 - 1/2), which is equal to 1/1. This is expressed as "1 to 1 odds" and often written "1:1".
Odds a:b for some event are equivalent to probability a/(a+b). For example, 1:1 odds are equivalent to probability 1/2, and 3:2 odds are equivalent to probability 3/5.
There remains the question of exactly what can be assigned probability, and how the numbers so assigned can be used; this is the question of probability interpretations. There are some who claim that probability can be assigned to any kind of an uncertain logical proposition; this is the Bayesian interpretation. There are others who argue that probability is properly applied only to propositions concerning sequences of repeated experiments or sampling from a large population; this is the frequentist interpretation. There are several other interpretations which are variations on one or the other of those, or which have less acceptance at present.
### Distributions
A probability distribution is a function that assigns probabilities to events or propositions. For any set of events or propositions there are many ways to assign probabilities, so the choice of one distribution or another is equivalent to making different assumptions about the events or propositions in question.
There are several equivalent ways to specify a probability distribution. Perhaps the most common is to specify a probability density function. Then the probability of an event or proposition is obtained by integrating the density function. The distribution function may also be specified directly. In one dimension, the distribution function is called the cumulative distribution function. Probability distributions can also be specified via moments or the characteristic function, or in still other ways.
A distribution is called a discrete distribution if it is defined on a countable, discrete set, such as a subset of the integers. A distribution is called a continuous distribution if it has a continuous distribution function, such as a polynomial or exponential function. Most distributions of practical importance are either discrete or continuous, but there are examples of distributions which are neither.
Important discrete distributions include the discrete uniform distribution, the Poisson distribution, the binomial distribution, the negative binomial distribution and the Maxwell-Boltzmann distribution.
Important continuous distributions include the normal distribution, the gamma distribution, the Student's t-distribution, and the exponential distribution.
## Probability in mathematics
Probability axioms form the basis for mathematical probability theory. Calculation of probabilities can often be determined using combinatorics or by applying the axioms directly. Probability applications include even more than statistics, which is usually based on the idea of probability distributions and the central limit theorem.
To give a mathematical meaning to probability, consider flipping a "fair" coin. Intuitively, the probability that heads will come up on any given coin toss is "obviously" 50%; but this statement alone lacks mathematical rigor - certainly, while we might expect that flipping such a coin 10 times will yield 5 heads and 5 tails, there is no guarantee that this will occur; it is possible for example to flip 10 heads in a row. What then does the number "50%" mean in this context?
One approach is to use the law of large numbers. In this case, we assume that we can perform any number of coin flips, with each coin flip being independent - that is to say, the outcome of each coin flip is unaffected by previous coin flips. If we perform N trials (coin flips), and let NH be the number of times the coin lands heads, then we can, for any N, consider the ratio NH/N.
As N gets larger and larger, we expect that in our example the ratio NH/N will get closer and closer to 1/2. This allows us to "define" the probability Pr(H) of flipping heads as the limit (mathematics), as N approaches infinity, of this sequence of ratios:
$\Pr(H) = \lim_{N \to \infty}{N_H \over N}$
In actual practice, of course, we cannot flip a coin an infinite number of times; so in general, this formula most accurately applies to situations in which we have already assigned an a priori probability to a particular outcome (in this case, our assumption that the coin was a "fair" coin). The law of large numbers then says that, given Pr(H), and any arbitrarily small number ε, there exists some number n such that for all N > n,
$\left| \Pr(H) - {N_H \over N}\right| < \epsilon$
In other words, by saying that "the probability of heads is 1/2", we mean that, if we flip our coin often enough, eventually the number of heads over the number of total flips will become arbitrarily close to 1/2; and will then stay at least as close to 1/2 for as long as we keep performing additional coin flips.
Note that a proper definition requires measure theory which provides means to cancel out those cases where the above limit does not provide the "right" result or is even undefined by showing that those cases have a measure of zero.
The a priori aspect of this approach to probability is sometimes troubling when applied to real world situations. For example, in the play Rosencrantz and Guildenstern are Dead by Tom Stoppard, a character flips a coin which keeps coming up heads over and over again, a hundred times. He can't decide whether this is just a random event - after all, it is possible (although unlikely) that a fair coin would give this result - or whether his assumption that the coin is fair is at fault.
### Remarks on probability calculations
The difficulty of probability calculations lie in determining the number of possible events, counting the occurrences of each event, counting the total number of possible events. Especially difficult is drawing meaningful conclusions from the probabilities calculated. An amusing probability riddle, the Monty Hall problem demonstrates the pitfalls nicely.
To learn more about the basics of probability theory, see the article on probability axioms and the article on Bayes' theorem that explains the use of conditional probabilities in case where the occurrence of two events is related.
## Applications of probability theory to everyday life
A major effect of probability theory on everyday life is in risk assessment and in trade on commodity markets. Governments typically apply probability methods in environment regulation where it is called "pathway analysis ", and are often measuring well-being using methods that are stochastic in nature, and choosing projects to undertake based on their perceived probable effect on the population as a whole, statistically. It is not correct to say that statistics are involved in the modelling itself, as typically the assessments of risk are one-time and thus require more fundamental probability models, e.g. "the probability of another 9/11". A law of small numbers tends to apply to all such choices and perception of the effect of such choices, which makes probability measures a political matter.
A good example is the effect of the perceived probability of any widespread Middle East conflict on oil prices - which have ripple effects in the economy as a whole. An assessment by a commodity trade that a war is more likely vs. less likely sends prices up or down, and signals other traders of that opinion. Accordingly, the probabilities are not assessed independently nor necessarily very rationally. The theory of behavioral finance emerged to describe the effect of such groupthink on pricing, on policy, and on peace and conflict.
It can reasonably be said that the discovery of rigorous methods to assess and combine probability assessments has had a profound effect on modern society. A good example is the application of game theory, itself based strictly on probability, to the Cold War and the mutual assured destruction doctrine. Accordingly, it may be of some importance to most citizens to understand how odds and probability assessments are made, and how they contribute to reputations and to decisions, especially in a democracy.
Another significant application of probability theory in everyday life is reliability. Many consumer products, such as automobiles and consumer electronics, utilize reliability theory in the design of the product in order to reduce the probability of failure. The probability of failure is also closely associated with the product's warranty.
## External links
• A Collection of articles on Probability, many of which are accompanied by Java simulations
• Edwin Thompson Jaynes. Probability Theory: The Logic of Science. Preprint: Washington University, (1996). -- HTML and PDF
• An online probability textbook which uses computer programming as a teaching aid
• Probabilistic football prediction competition, probabilistic scoring and further reading.
• "The Not So Random Coin Toss, Mathematicians Say Slight but Real Bias Toward Heads". NPR.
• Figuring the Odds (Probability Puzzles)
• Dictionary of the History of Ideas: Certainty in Seventeenth-Century Thought
• Dictionary of the History of Ideas: Certainty since the Seventeenth Century
## Quotations
• Damon Runyon, "It may be that the race is not always to the swift, nor the battle to the strong - but that is the way to bet."
• Pierre-Simon Laplace "It is remarkable that a science which began with the consideration of games of chance should have become the most important object of human knowledge." Théorie Analytique des Probabilités, 1812.
• Richard von Mises "The unlimited extension of the validity of the exact sciences was a characteristic feature of the exaggerated rationalism of the eighteenth century" (in reference to Laplace). Probability, Statistics, and Truth, p 9. Dover edition, 1981 (republication of second English edition, 1957).
03-10-2013 05:06:04
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http://mathhelpforum.com/discrete-math/199986-set-open.html
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1Thanks
• 1 Post By mfb
# Thread:
1. ## Is this set open?
A is a subset of B, both sets are open.
Is the set B/A open?
My answer would be no, because the complement of an open set is always closed. Is this the correct definition in this case?
2. ## Re: Is this set open?
But a set can be both open and closed.
You can look at A=(0,1), B=(0,2) as subset of the real line with the usual topology.
3. ## Re: Is this set open?
girdav: In your example, A\B is [1,2) which is not open.
However, let B be the union of two arbitrary disjoint open subsets A and C. B is open.
Then B\A = C is open by construction.
4. ## Re: Is this set open?
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
5. ## Re: Is this set open?
Originally Posted by infernalmich
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
You must understand that $B\setminus A=B\cap A^c$ and is known as a relative complement.
That is the complement of A relative to B.
You see $A$ is an open set in the underlying space which may not be $B$.
In the real numbers $((0,2)\setminus (0,1)=[1,2)$ which is not open.
6. ## Re: Is this set open?
Originally Posted by infernalmich
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
Yes, that's true. But "B\A" is NOT the "complement" of A, which is, by definition, the set all points not in A, whether they are in B or not. If $B= (0, 1)\cup (2, 3)$ and $A= (0, 1)$, then $B\A= (2, 3)$ an open set. The [b]complement of A is $(-\infty, 0]\cup [1, \infty$, a closed set.
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http://math.stackexchange.com/questions/160662/what-transforms-under-su2-as-a-matrix-under-so3
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# What transforms under SU(2) as a matrix under SO(3)?
A vector $\boldsymbol{r}$ in $\mathbb{R}^3$ transforms under rotation $\boldsymbol{A}$ to $\boldsymbol{r}'=\boldsymbol{Ar}$. It is equivalent to an SU(2) "rotation" as $$\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdot\boldsymbol{\sigma} \right) \boldsymbol{h}^{-1},$$ where $\boldsymbol{h}$ is the counterpart of $\boldsymbol{A}$ in SU(2) given by the homomorphism between these two groups.
Now the question is, what would be the equivalent transformation in SU(2) of the rotation of a matrix in $\mathbb{R}^3$? In other words, what is the equivalent in SU(2) of $\boldsymbol{M}'=\boldsymbol{A}\boldsymbol{M}\boldsymbol{A}^{-1}$.
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3
I am a little confused about what you're asking. You presumably know that there is a double cover $\phi : \text{SU}(2) \to \text{SO}(3)$. Given any action of $\text{SO}(3)$ on any kind of thing, pulling back along $\phi$ gives you an action of $\text{SU}(2)$ on that thing (an element $g \in \text{SU}(2)$ acts by however $\phi(g)$ acts). – Qiaochu Yuan Jun 20 '12 at 7:15
@QiaochuYuan: I didn't learn the concept of "pullback". Can you elaborate? – C.R. Jun 20 '12 at 7:33
1
All I mean is that if $S$ is a set (e.g. a vector space) and $\rho : \text{SO}(3) \to \text{Aut}(S)$ is an action of $\text{SO}(3)$ on that set (e.g. a linear representation on a vector space) then $\rho \circ \phi : \text{SU}(2) \to \text{Aut}(S)$ is the corresponding action of $\text{SU}(2)$. – Qiaochu Yuan Jun 20 '12 at 8:11
3
@Qiaochu: I think the OP is asking for an explicit formula for $\rho \circ \phi$ where $S = \mathcal{M}_{3\times 3}$ and $\rho(A)M = AMA^{-1}$. – Willie Wong♦ Jun 20 '12 at 8:15
## 1 Answer
Firstly, we need to map $\mathbb{R}^3$ to the representation space $V$ for $\mathrm{SU}(2)$. One possible map is given by the following formula: $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto x \mathbf{I} + y \mathbf{J} + z \mathbf{K}$$ where \begin{align} \mathbf{I} & = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} & \mathbf{J} & = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} & \mathbf{K} & = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \end{align}
$\mathrm{SU}(2)$ acts on $V$ by conjugation: so for each $X$ in $V$ and each $A$ in $\mathrm{SU}(2)$, the ordinary matrix product $A X A^{-1}$ is in $V$. This is linear in $X$ and is indeed a linear representation of $\mathrm{SU}(2)$. Indeed, if $$A = \begin{pmatrix} r e^{i \theta} & s e^{-i \phi} \\ -s e^{i \phi} & r e^{-i \theta} \end{pmatrix}$$ where $r, s, \theta, \phi$ are real numbers and $r^2 + s^2 = 1$, then $A \in \mathrm{SU}(2)$, and \begin{align} A \mathbf{I} A^{-1} & = (r^2 - s^2) \mathbf{I} + 2 r s \sin (\theta - \phi) \mathbf{J} - 2 r s \cos (\theta - \phi) \mathbf{K} \\ A \mathbf{J} A^{-1} & = 2 r s \sin (\theta + \phi) \mathbf{I} + (r^2 \cos 2 \theta + s^2 \cos 2 \phi) \mathbf{J} + (r^2 \sin 2 \theta - s^2 \sin 2 \phi) \mathbf{K} \\ A \mathbf{K} A^{-1} & = 2 r s \cos (\theta + \phi) \mathbf{I} - (r^2 \sin 2 \theta + s^2 \sin 2 \phi) \mathbf{J} + (r^2 \cos 2 \theta - s^2 \cos 2 \phi) \mathbf{K} \end{align} Thus, the induced action of $\mathrm{SU}(2)$ on $\mathbb{R}^3$ is given by the group homomorphism below, $$\begin{pmatrix} r e^{i \theta} & s e^{-i \phi} \\ -s e^{i \phi} & r e^{-i \theta} \end{pmatrix} \mapsto \begin{pmatrix} r^2 - s^2 & 2 r s \sin (\theta + \phi) & 2 r s \cos (\theta + \phi) \\ 2 r s \sin (\theta - \phi) & r^2 \cos 2 \theta + s^2 \cos 2 \phi & -r^2 \sin 2 \theta - s^2 \sin 2 \phi \\ -2 r s \cos (\theta - \phi) & r^2 \sin 2 \theta - s^2 \sin 2 \phi & r^2 \cos 2 \theta - s^2 \cos 2 \phi \end{pmatrix}$$ and one may verify that the RHS is a matrix in $\mathrm{SO}(3)$.
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You are just re-expressing what I already know, that $\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdot\boldsymbol{\sigma} \right) \boldsymbol{h}^{-1}$. – C.R. Jun 20 '12 at 10:47
@KarsusRen: Then what would you like to know? – Zhen Lin Jun 20 '12 at 11:14
I stated clearly in my question: what is the equivalent transformation in SU(2) of the rotation of a linear operator (matrix) instead of vector in $\mathbb{R}^3$? – C.R. Jun 20 '12 at 11:54
Just conjugate by the corresponding $\mathrm{SO}(3)$ matrix. There isn't a nice way of representing it as a matrix formula in terms of the $\mathrm{SU}(2)$ matrix. (Think about it: if a vector in $\mathbb{R}^3$ corresponds to a matrix in $V$, then a matrix would have to correspond to some hideous rank-3 tensor!) – Zhen Lin Jun 20 '12 at 13:22
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http://mathhelpforum.com/calculus/68552-cauchy-schwarz-inequality-integrals.html
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# Thread:
1. ## Cauchy-Schwarz inequality for Integrals
Suppose that the functions $f,g,f^2,g^2, fg$ are integrable on the closed, bounded interval [a,b]. Prove that:
$\int ^b _a fg \leq \sqrt { \int ^b_a f^2 } \sqrt { \int ^b_a g^2 }$.
Proof.
Now, $0 \leq \sqrt { \int _a ^b (f- \lambda g)(f- \lambda g) } ^2$ $= \int ^b_a (f^2- \lambda fg - \lambda fg + \lambda ^2 g^2 )$
$= \int ^b_a f^2 - \lambda \int ^b _a fg - \lambda \int ^b _a fg + \lambda ^2 \int ^b _a g^2$
Choose $\lambda = \int _a ^b fg [ \int ^b _a g^ {2}]^{-1}$
then the previous equation becomes $\int ^b _a f^2 - \int ^b _a (fg)^2 [ \int ^b _a g^2 ]^{-1} - \int ^b _a (fg)^2 [ \int ^b _a g^2 ] ^{-1} + \int ^b _a (fg)^2$
which implies $0 \leq \int ^b _a f^2 - \int ^b _a (fg)^2 (2 [ \int ^b _a g^2 ]^{-1})$
$\int ^b _a (fg)^2 (2 [ \int ^b _a g^2 ] ^{-1} \leq \int ^b _a f^2$
$\int ^b _a (fg)^2 \leq \frac { \int ^b _a f^2 \int ^b _a g^2 } {2}$
How do I get rid of the 2 at the bottom?
Thanks.
2. The Cauchy-Schwarz inequality is a general inequality for inner product spaces.
Thus, what you wrote is a consequence of this general inequality found here*.
*)I am just a little worried because the space of all integral functions is not an inner product space with norm $\left< f,g\right> = \smallint_a^b fg$.
However, the proof on that page seems to work.
3. Hello,
A more general view on this inequality is Hölder's inequality.
Cauchy-Schwarz is then a specific situation, with p=q=2.
Hölder's inequality - Wikipedia, the free encyclopedia
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http://mathoverflow.net/questions/116701?sort=newest
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## How would set theory research be affected by using ETCS instead of ZFC?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In "Rethinking Set Theory", Tom Leinster argues in favor of teaching axiomatic set theory via Lawvere's Elementary Theory of the Category of Sets with 10 axioms (but phrased in a way that requires no knowledge of category theory) which uses sets and functions as primitive elements compared to ZFC which uses sets and elements as primitives. (although if you really insist you can reintroduce elements as primitives at the expense of more clauses). This is weaker than ZFC but can be made equivalent to ZFC (or equiconsistent according to François G. Dorais) with the addition of a hardly ever needed 11th axiom.
http://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html
How would set theory research be affected by using ETCS instead of ZFC?
Is ETCS less cumbersome than ZFC or more, or does it not matter? Does it make proofs longer/shorter or easier/harder?
Would automated theorem provers work better with ETCS?
To clarify, the question is not about the strength of ETCS: assume whatever is needed to bring ETCS up to strength with ZFC. The question is about practicalities of working with the ETCS axioms. I guess most mathematicians don't directly work with axioms of set-theory at all. So the question is: if your work does involve working directly with such axioms then what difference (if any) does it practically make to you as a set-theory researcher to use a different set of axioms that have equal strength. If I understand correctly the goal of Tom Leinster's paper was to reflect actual (non-set-theoretical)-mathematical practice but my question is about what set-theorists would do if they used these axioms. Assume also that for the purposes of this question ETCS refers specifically to the rephrasing used in this paper that is free of category and topos terminology. Perhaps this rephrasing should be renamed ETS. Adding replacement would give the name ETSR.
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7
This seems a bit vague as an MO question (seen in its totally, parts might not be). In addition, this blog post was made today (or yesterday) and there is some discussion in the comments there. Why don't you participate in it or at least wait until that disucssion is over? (I did not yet vote to close but am considering it.) – quid Dec 18 at 14:19
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Martin, what's not to love about the current definition of ordinals? It's wunderbar! – Asaf Karagila Dec 19 at 17:42
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It's debatable that ETCS reflects actual non-set-theoretic mathematical practice, since it has a very restricted notion of object identity. For example, two sets don't share elements, since two functions with different codomains are not equal. The number 2 as a membrer of the natural numbers and the number 2 in the even numbers are not equal. There are standard ways around this, so maybe it's not important, but it is a departure from standard mathematical practice. – arsmath Jan 6 at 22:20
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@arsmath: Even in ZFC, the natural number 2 is not the same as the real number 2; the former is usually the set $\{\emptyset,\{\emptyset\}\}$ while the latter is something like the set of all rational numbers less than 2. I think this is best considered an issue of "implicit coercions" which can even be made precise in a computer proof assistant; it's not really something specific to ZFC or ETCS. – Mike Shulman Jan 7 at 7:28
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Mike Shulman has written new blog post "From Set Theory to Type Theory" golem.ph.utexas.edu/category/2013/01/… "My goal for this post is to start from material set theories (like ZFC) and structural set theories (like ETCS), and show how type theory can arise naturally out of an attempt to take the best of both worlds. At the end, I’ll argue that to really resolve all the problems, we need univalent type theory." – unknown (yahoo) Jan 7 at 14:04
show 13 more comments
## 7 Answers
No one has really addressed the question about automated theorem provers. I'm going to take this to refer more generally to computer-assisted mathematics, since fully automated theorem provers are still of very limited usefulness to a mathematician. There are a lot of good arguments that the best foundational system for these is neither ZFC (or its relatives) nor ETCS (or its relatives), but some variety of type theory.
This is mainly because type theory is so closely connected with programming and has good computation behavior, thus meshing very well with computers. Since the question is not about type theory, I won't say much about it; I just want to point out that ETCS is more like type theory than ZFC is. So although I kind of doubt that actually using ETCS in a computer proof assistant would be much better than using ZFC, learning to think in ETCS and do mathematics therein will probably stand you in better stead when you try to use a proof assistant based on type theory.
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Isabelle/HOL is a concrete example of a proof assistant using type theory. I completely agree that ETCS is easier to understand as variety of type theory, in the sense that ETCS regards each set as its own type, and the "element" relation as a fact about the type of the element. However, I'm not yet convinced that learning to do mathematics in ETCS specifically is likely to help with proof assistant use. Natural-language math is done in a complex type theory already, so most mathematicians are accustomed to thinking this way. Learning formal type theory might be a better preparation. – Carl Mummert Jan 7 at 13:15
Learning formal type theory would undoubtedly be a better preparation! – Mike Shulman Jan 9 at 9:22
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Some quotes:
"I hope the math community has reached the point of realizing that we really need not one foundation of mathematics, but many, together with clearly described relations between them. Indeed at this point the word ‘foundation’ is perhaps less helpful than something else… like maybe ‘entrance’." , John Baez http://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html#c042716
"There are some problems with this sort of compromise where material and structural set theories cohabitate as equal alternative foundations. ... this kind of cohabitation model is actually not the best possible way to accommodate both the material and the structural perspective at the foundational level. ... what we end up doing is moving all the annoyances from both sides into the translation between the two ... it would be much more satisfactory to have a greater overarching foundation similar to Cantor’s universe that comprises both the material and structural views.", François G. Dorais http://dorais.org/archives/1135
"if you care about internalizing mathematics in nonstandard models — not only in order to prove formal independence results, as set theorists do, but because you care about the nonstandard models in their own right and you want to use internalization to prove things about them — then ETCS is your friend and ZFC is your enemy. Structural set theory — and its sister, type theory — is very well adapted to internalizing in many different contexts" , Mike Shulman http://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html#c042933
"with ZFC is that it allows things like ⟨0,1⟩⊆3 which make no sense mathematically. I can assure you, though, that after you write one piece of code in x86 assembler which rewrites itself as it runs that manipulating all objects as binary and manipulating all objects as sets make perfect sense. This is what annoys me, in some sense, that people have an issue that in a system which allows you to encode everything as sets you can write some weird junk. Equivalently, this would be baffling to those people that if you write a piece of code in Common Lisp then the code itself is just a list and the program can manipulate itself (easier than to do that in assembler, I have to admit).", Asaf Karagila http://boolesrings.org/asafk/2013/on-leinsters-rethinking-set-theory/#comment-789
"I don’t like the hard line that certain things are meaningless, simply because this depends on the setting. I don’t object to the idea of saying that an expression is, for your current purposes, well-typed, since this just means you say nothing about non-well-typed expressions. I would say not “meaningless”, but rather “does not have an intended meaning”. There are expressions, such as 0∈1 that do not have an intended meaning. If they have a meaning in a certain setting, then so be it.", Walt http://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html#c042992
"ETCS is one, but apparently not one that is very congenial to many traditional set theorists, for a variety of reasons, maybe one being that many set theorists “don’t like categories” (well, okay, but again I think that’s a pity). Maybe some set theorists don’t like the privileging of functions over elements (although one take is that functions are generalized elements).", Todd Trimble, http://boolesrings.org/asafk/2013/on-leinsters-rethinking-set-theory/#comment-798
"The same arrows aimed at ZFC can be easily turned at ETCS. It is equally easy to ask about “senseless things” which happen in ETCS but not in general mathematics. And this is not a critique against ETCS, but rather against the critique aimed at ZFC.", Asaf Karagila http://boolesrings.org/asafk/2013/on-leinsters-rethinking-set-theory/#comment-815
"the issue is that first-order logic is not “type safe”, and I would suspect that people who have had exposure to first-order logic take this for granted, while those who have not may find it surprising. The reason is that syntax, in first-order logic, is defined independently of semantics, so that whether a formula is well-formed (syntactically correct) is independent of what the formula means. ... This issue of “type safety” arises also in programming languages. Some languages, such as C, have little run-time type checking. I can take a 64 bit integer and treat it as an array of eight 8-bit characters – with implementation-defined results. ... In working mathematics we have humans to check that the formulas we actually write make sense – nobody accidentally writes 1∈sin in an analysis paper – so it is not clear that there is any real advantage to stronger “type checking” in the foundations.", Carl Mummert http://boolesrings.org/asafk/2013/on-leinsters-rethinking-set-theory/#comment-794
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Dear Unknown, I think this should be a community wiki since others might want to add to this interesting sequence of quotes. (You can find the community wiki checkbox at the bottom right of the edit box.) – François G. Dorais♦ Jan 6 at 22:23
Good idea. Done. – unknown (yahoo) Jan 6 at 22:31
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I have to admit a certain excitation to see quotes from the comments on my blog. Huzzah. :-) – Asaf Karagila Jan 6 at 22:35
In principle, nothing would be lost by working in ETCS+R instead of ZFC since the two theories are very nicely interpretable in each other. However, after talking it through with Mike Shulman (whom I thank very much) here, I came to the conclusion that much would be lost in practice.
The problem lies with the notion of wellfoundedness. Recall that a binary relation $R$ on a set $A$ is wellfounded if every nonempty subset $B$ of $A$ has an $R$-minimal element (an $x \in B$ such that $y \not\mathrel{R} x$ for all $y \in B$); this is a $\Pi_1$ statement in the Lévy hierarchy. In ZFC, an equivalent statement is that there is an ordinal valued rank function for $R$ on $A$ (a function $r:A \to \mathrm{Ord}$ such that $x \mathrel{R} y \Rightarrow f(x) \lt f(y)$ for all $x,y\in A$). This is a $\Sigma_1$ statement in the Lévy hierarchy due to the fact that being an ordinal is $\Delta_0$. In ETCS+R, there is no suitable equivalent to ordinals: two wellorderings of the same type are indistinguishable from each other. Because of this, in ETCS+R, the equivalent rank function statement is much more complex since one must explicitly say that the codomain of the rank function is a wellordering (for which we must use the $\Pi_1$ definition).
Since wellfoundedness is $\Delta_1$ in ZFC, it is absolute for transitive models (of a small fragment) of ZFC. In other words, transitive embeddings between models of ZFC preserve wellfoundedness. (An embedding $f:M \to N$ is transitive if $f$ maps the elements of $x$ in $M$ onto the elements of $f(x)$ in $N$ for every $x \in M$; so the range of $f$ is a transitive substructure of $N$ isomorphic to $M$.) For models of ETCS+R, there is no natural equivalent to transitive embeddings so if one wants to preserve wellfoundedness then one must explicitly require that. In practice, it is very difficult to check that an embedding preserves wellfoundedness, especially when compared to checking that an embedding is transitive.
Therefore, ETCS+R does not have a very good grasp of wellfoundedness compared to ZFC. Since the study of wellfoundedness is so central to modern set theory, the framework of ZFC is much more appropriate than ETCS+R for set theorists to work with. To rephrase an analogy I used elsewhere, asking a set theorist to work in ETCS+R instead of ZFC is like asking a ring theorist to work with ternary relations $A(x,y,z)$ for addition $x+y = z$ and $M(x,y,z)$ for multiplication $x \cdot y = z$: it's essentially the same, in principle, but it's simply not the right thing to do!
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Hear hear! In September I worked a bit with Misha Gavrilovich on his model-categorical construction QtNaamen, and I realized that diagrams are simply not the right language to talk about models of ZFC, and in particular when you want to talk about elementary embeddings or so. There are some semi-positive conjectures, and we came up with an idea on how to fine tune it to allow the elementarity to slip through the cracks, but it's still very much unfit as a language. (I read your blog post and liked it, and it reminded me to write a post on the work with Misha...) – Asaf Karagila Jan 4 at 1:27
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For the record, let me repeat here my essential agreement, and also my linguistic objection. I would say the problem isn't that ETCS has a "weak grasp of well-foundedness; ETCS understands well-founded relations perfectly well as a structural concept. The problem is more like that in ETCS the "universe" is not well-founded in any sense: ZFC only knows about well-founded things, while ETCS knows about other things as well, and that makes it more complicated to use if all you care about are the well-founded ones. – Mike Shulman Jan 7 at 7:32
There are several relatives (typically: subtheories and supertheories) of ZFC that are used in set-theoretic research. If somebody wants to do set theory based on ETCS or a related base theory, these theories would have to be translated to this new base (or, in the case of theories that are not so canonical, such as ZFC*, substitutes would have to be found).
Such a translation seems to be rather straightforward in the case of supertheories; a good translation would of course use the idioms of the target language.
• Examples of supertheories:
• ZFC plus large cardinals
• ZFC plus definable determinacy (e.g., projective determinacy, or some consequences thereof - related to large cardinals).
• ZF plus AD, or ZF + $V=L(\mathbb R)$ (supertheory of ZF only)
• ZF(C) plus "V is a certain inner model". Weaker versions include:
• ZFC plus cardinal arithmetic assumptions (GCH, SCH)
• ZFC plus combinatorial principles ($\diamondsuit$, etc)
• ZFC plus forcing axioms (MA, PFA etc)
• others, including combinations (conjunctions) of the above
• Examples of subtheories
• KP and related theories, which do not have full comprehension. (I think they are usually associated with proof theory rather than set theory)
• ZFC minus infinity (this is more closely associated with arithmetic than with set theory)
• ZF, or ZF plus weak versions of choice
• ZFC minus power set (plus instances of power set). Typical models are of the form $H(\chi)$.
• ZFC minus replacement (plus finitely many instance of replacement). Typical models are $V_\delta$.
• ZFC*, an often unspecified finite subset of ZFC, used to get around the "undefinability of truth", or to apply the reflection theorem. (Morally the same as the previous item.)
• ZFC minus Foundation, reflecting the fact that Foundation is hardly used outside set theory.
• Others, including combinations (intersections) of the above
• Other relatives:
• ZF(C) with atoms, perhaps closer to mathematical practice than ZFC itself.
• NBG. The relation between NBG and ZFC is very well understood, as are the advantages and disadvantages: On the plus side, NBG can naturally talk about classes, and is finitely axiomatized (which might make it more amenable to automated theorem proving). On the other hand, the fact that not every subclass of $\mathbb N$ is a set can be inconvenient.
• MK and others.
• NF and NFU, sometimes claimed to be more natural than ZFC. While ZFC makes it awkward to talk about classes, NFU has problems with the function `$x\mapsto \{x\}$`.
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Is there a simple example of a subclass of $\omega$ which is not a set? (I suspect it will be something related to undefinability of the truth or so?) – Asaf Karagila Jan 4 at 1:49
Unless I'm confused (which is possible), there is no refutation in NBG of the claim that every subclass of a set X is a set; there just is no proof (not even if we require X to be ω, although there is if we require X to be finite). If you add this claim (as an axiom scheme, and then the result is no longer finitely axiomatisable) to NBG, then you get the consistent (assuming that ZFC plus one inaccessible cardinal, or even something weaker than that, is consistent) theory MK (Morse–Kelley). – Toby Bartels Jan 7 at 9:56
Well, one would just add a just strong enough instance of replacement, or the consequence one actually needs, to the assumptions of the theorem one was proving. For instance, one could take the existence of $\aleph_\omega$ as an assumption (like a large cardinal) rather than proving it exists. There are models of ETCS in which this exists and in which it doesn't. In this instance, one is just lowering the level at which one considers a cardinal 'large'.
I'm reminded of McLarty's recent work on weak set theories, and doing algebraic geometry in them. One can prove strong results about cohomology, but can't prove the existence of an uncountable ring like $\mathbb{R}$ or $\mathbb{C}$. He says you can just simply posit the existence of such a thing on top of the weak set theory, and then all the theorems go through.
Of course, in set theory there are a whole stack of results that rely on the strength of ZFC beyond ETCS, and it would entail a whole lot of reverse mathematical effort to find out what is actually necessary for swathes of theory, that many would say it is easier to just assume what ZFC gives you. But it isn't strictly necessary, just a lot easier.
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I have to admit often that I am weirded out by all those theories which are like "We are better than ZF, but we can't prove anything useful exists. But if you assert its existence then the theorems work as before". Wasn't the idea of foundations on set theory exactly to have one framework which just proves that everything you want to do can be done, except division by zero? :-) – Asaf Karagila Dec 18 at 23:35
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... just a foundation for 'ordinary/everyday' maths. – David Roberts Dec 19 at 0:08
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@Asaf: although I am not sure what question David is responding to in the OP (see also my comment under Andreas's answer), your comment sounds just a wee bit aggressive to me. Maybe the question needs to be closed down, before this gets too subjective and argumentative? – Todd Trimble Dec 19 at 0:13
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My reading of Asaf's comment is different than Todd and Andrej. I think this is a case where the medium of communication interferes with the message. I think his (poorly formulated) point is valid. It's not clear to me that a general foundation should have the goal of doing just enough instead of the goal of doing it all. That said, there are other contexts where doing the minimum makes perfect sense, such as reverse mathematics. – François G. Dorais♦ Dec 19 at 1:54
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Well, I do feel we're getting sidetracked. For the sake of making progress, how about we consider ETCS augmented by a suitable axiom scheme of replacement (a categorical form of this has been discussed by McLarty -- the reference is in Tom's paper, IIRC), and try addressing OP's questions with that in mind? – Todd Trimble Dec 19 at 2:57
show 4 more comments
I like the fact that you asked this question, but I'm a little worried that it will be seen as "subjective and argumentative".
For the second question, I think a lot of set theorists (trained since childhood in ZFC) have gone on record saying they do find ETCS not easy to work with or not user-friendly, and I find this quite understandable. There is a critical hump of lemmas to get over in the beginning when working with ETCS, because unlike ZFC or some variation thereof, there is no ready-made comprehension scheme in ETCS (at least as it is ordinarily presented). Rather, a stock of instances of comprehension (where one builds up an "internal logic" by hand, as it were) have to proven before one is ready to fly. But after some point, with enough of these beginning lemmas under the belt, the development of mathematics, say of the core undergraduate curriculum involving basic results of real analysis, algebra, topology, etc., proceeds pretty much the same way as what one is used to. So in that sense, I'd say "it does not matter" for the needs of working mathematicians.
In developing his new set theory SEAR (Sets, Elements, And Relations), Mike Shulman recognized this cumbersome aspect of ETCS, quoting an analogy I made once about ETCS on the now-moribund "Todd and Vishal's blog":
[Trimble] "with ZFC it’s more as if you can just hop in the car and go; with ETCS you build the car engine from smaller parts with your bare hands, but in the process you become an expert mechanic, and are not so rigidly attached to a particular make and model."
[Shulman] Using this metaphor, SEAR can be thought of as an ETCS-car which comes preassembled with a nice slick control panel. Or, using an alternate metaphor, ZFC is like Windows, ETCS is like UNIX, and SEAR is like OS X (or maybe Ubuntu). With SEAR you get a nice familiar interface with which it is easy to do standard things, there is less cruft than you get with ZFC, and behind the scenes you have all the power of ETCS (and more). (Of course, if you like Microsoft products, then this metaphor probably does not appeal to you.)
So SEAR aims to combine the advantages of ETCS (as a truly "structural" set theory, where the aspects of sets we truly care about are isomorphism-invariant, and carry less "cruft") with the advantages of ZFC (where comprehension principles, etc. are built right in; see his axiom 1). SEAR will probably look more home-y and familiar to those used to traditional naive set theory; there is none of this intimidating "well-pointed topos with NNO and choice" business to wade through at the outset.
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@Todd Trimble: This answer is closest to the intention of the question, particularly with phrases like "gone on record saying they do find ETCS not easy to work with or not user-friendly". – Stxmqs Dec 19 at 6:54
Much of set theory research (for example, a paper I'm currently working on, concerning certain ultrafilters over the natural numbers) would work just fine in ETCS. But much of set theory research uses the axiom of replacement in ways that cannot be imitated in ETCS. Tom Leinster's paper gives the example of the existence of the union of $\mathbb N$, its power set $\mathcal P(\mathbb N)$, the power set of that, and so on, iterated for countably many steps. To put it another way, ETCS cannot prove that there is a cardinal number $\kappa$ with infinitely many infinite cardinals below it. Such $\kappa$'s and far larger ones are certainly involved in a great deal of research-level set theory, so a foundation having the strength of ZFC (or even more) is needed there.
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My impression is that the OP was aware of the need for an extra replacement axiom in addition to ETCS to get equiconsistency with full ZFC; his questions seem to lie elsewhere. – Todd Trimble Dec 19 at 0:06
@Andreas Blass: Ok, so set-theory goes well beyond ZFC these days. The question then is take ETCS as your base and add whatever large cardinals that you need, then do you now care that you have started with ETCS instead of ZFC? – Stxmqs Dec 19 at 6:48
Another question arises: can even stronger theories be rephrased using ideas that would be comfortable for category-theorists but without the category terminology - category-free categorical-set theories and would set-theorists care for such things? – Stxmqs Dec 19 at 7:01
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As Tom Leinster points out in his paper, one can reconstruct the ZFC universe in ETCS once one adds enough axioms (like a version of replacement). After that reconstruction is carried out, there would be no noticeable difference between working in ZFC and working in the interpretation of ZFC in ETCS. You could take what people ordinarily say on the basis of ZFC, including things about the ZFC membership relation and the cumulative hierarchy of sets, and understand them as referring to those notions as defined within ETCS. – Andreas Blass Dec 19 at 12:38
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http://www.physicsforums.com/showthread.php?t=260849
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Physics Forums
Right handed vs Left handed circular polarization
Hey,
I just wanted to clear up some confusion I've been having regarded which is which of these.
If I have the wave $$\vec{E}= E_{0X} cos(kz-\omega t)+ E_{0Y} sin(kz-\omega t)$$ and $$E_{0X}=E_{0Y}$$. Then at z=0, t=0 the field is pointing completely in the x direction. Staying at z=0 ( $$\vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)-E_{0Y}sin(\omega t)$$. So that the x comp begins to decrease with time at z=0, and the y comp begins to increase negativley. Thus if the wave was coming toward you down the z-axis you'd see it rotating clockwise. If you were behind the wave you'd see it rotating counter clockwise.
This is right handed circ polarization?
Now if I have If I have the wave $$\vec{E}= E_{0X} cos(kz-\omega t)- E_{0Y} sin(kz-\omega t)$$ and $$E_{0X}=E_{0Y}$$. Then at z=0, t=0 the field is pointing completely in the x direction again. Staying at z=0 ( $$\vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)+E_{0Y}sin(\omega t)$$. So that the x comp begins to decrease with time at z=0, and the y comp begins to increase positivley. Thus if the wave was coming toward you down the z-axis you'd see it rotating counter-clockwise. If you were behind the wave you'd see it rotating clockwise.
This is left handed circ polarization?
Does all this sound correct, and are these the conventions?
Thanks
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Science Advisor You have the correct conventions for optics. In high energy physics, the point of view is that of the photon, so you do look behind the wave. This means that negative helicity (or left-handed helicity) corresponds to right handed polarization.
Quote by h0dgey84bc ... Thus if the wave was coming toward you down the z-axis you'd see it rotating clockwise. If you were behind the wave you'd see it rotating counter clockwise. This is right handed circ polarization?
Not per IEEE-STD-145.
$$E_{CP}=E_x \pm jE_y$$
Do you understand complex exponentials?
Regards,
Bill
Right handed vs Left handed circular polarization
Not per IEEE-STD-145. [tex]E_{CP}=E_x \pm jE_y[tex] Do you understand complex exponentials?
Sure. You're referring to the phasor representation?
As long as the convention I'm using with be good with the GRE I'm happy, just don't want to lose marks stupidly for the wrong convention.
Recognitions:
Science Advisor
Quote by Antenna Guy Not per IEEE-STD-145. $$E_{CP}=E_x \pm jE_y$$ Do you understand complex exponentials? Regards, Bill
The first post is just the trig representation of your exponentials.
The question was is the + or - for right handed circular polarization.
Quote by clem The first post is just the trig representation of your exponentials. The question was is the + or - for right handed circular polarization.
Note from phasors that a factor of $j$ is equivalent to a +90deg phase shift.
Assume that the phase convention is such that I am looking in the direction of propogation.
For RHCP, $E_y$ "lags" $E_x$, and a factor of j would rotate the phasor of $E_y$ onto that of $E_x$. Another way of looking at it is that after a quarter wavelength of propogation, $E_y$ would have the phase that $E_x$ started with.
Hence:
$$E_R=E_x+jE_y$$
$$E_L=E_x-jE_y$$
If $E_x$ and $E_y$ are co-phase, the total field is linear (i.e. non-rotating). In this case, $E_R$ and $E_L$ are simply conjugates of one-another (same magnitude).
If $E_x$ and $E_y$ differ by exactly 90deg of phase (and have the same magnitude), the total field is either $E_R$ or $E_L$.
In any other case the total field is elliptical, and has both $E_R$ and $E_L$ components.
Regards,
Bill
Recognitions: Science Advisor Bill: You "Assume that the phase convention is such that I am looking in the direction of propagation." The usual convention in optics is that you are looking at the light coming toward you. This is opposite to the direction of propagation. You are using the high energy convention for photons, which is fine, but not what the original questioner asked about.
Mentor
Blog Entries: 10
Quote by h0dgey84bc Hey, I just wanted to clear up some confusion I've been having regarded which is which of these. If I have the wave $$\vec{E}= E_{0X} cos(kz-\omega t)+ E_{0Y} sin(kz-\omega t)$$ and $$E_{0X}=E_{0Y}$$. Then at z=0, t=0 the field is pointing completely in the x direction. Staying at z=0 ( $$\vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)-E_{0Y}sin(\omega t)$$. So that the x comp begins to decrease with time at z=0, and the y comp begins to increase negativley. Thus if the wave was coming toward you down the z-axis you'd see it rotating clockwise. If you were behind the wave you'd see it rotating counter clockwise. This is right handed circ polarization?
Yes, it's right-handed. To see this let t=0. As you go in the +z direction, E rotates from +x to +y, then -x, then -y. Letting the fingers of your right hand curl around in the direction of rotation, your right thumb points in the +z direction. The picture is the same as the threading on a standard right-handed screw.
For your other case, the same argument works out if you use your left hand, so it is left-handed circular polarization.
You could also imagine going in the -z direction, these arguments would still work.
Quote by clem Bill: You "Assume that the phase convention is such that I am looking in the direction of propagation." The usual convention in optics is that you are looking at the light coming toward you. This is opposite to the direction of propagation. You are using the high energy convention for photons, which is fine, but not what the original questioner asked about.
The OP never said anything about "optics" (or high energy physics).
I cited an international standard, and (hopefully) followed it accurately.
Regards,
Bill
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http://unapologetic.wordpress.com/2009/06/
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# The Unapologetic Mathematician
## Dirac notation I
There’s a really neat notation for inner product spaces invented by Paul Dirac that physicists use all the time. It really brings to the fore the way a both slots of the inner product enter on an equal footing.
First, we have a bracket, which brings together two vectors
$\displaystyle\langle w,v\rangle$
the two sides of the product are almost the same, except that the first slot is antilinear — it takes the complex conjugate of scalar multiples — while the second one is linear. Still, we’ve got one antilinear vector variable, and one linear vector variable, and when we bring them together we get a scalar. The first change we’ll make is just to tweak that comma a bit
$\displaystyle\langle w\vert v\rangle$
Now it doesn’t look as much like a list of variables, but it suggests we pry this bracket apart at the seam
$\displaystyle\langle w\rvert\lvert v\rangle$
We’ve broken up the bracket into a “bra-ket”, composed of a “ket” vector $\lvert v\rangle$ and a “bra” dual vector $\langle w\rvert$ (pause here to let the giggling subside) (seriously, I taught this to middle- and high-schoolers once).
In this notation, we write vectors in $V$ as kets, with some signifier inside the ket symbol. Often this might be the name of the vector, as in $\lvert v\rangle$, but it can be anything that sufficiently identifies the vector. One common choice is to specify a basis that we would usually write $\left\{e_i\right\}$. But the index is sufficient to identify a basis vector, so we might write $\lvert1\rangle$, $\lvert2\rangle$, $\lvert3\rangle$, and so on to denote basis vectors. That is, $\lvert i\rangle=e_i$. We can even extend this idea into tensor products as follows
$\displaystyle e_i\otimes e_j=\lvert i\rangle\otimes\lvert j\rangle=\lvert i,j\rangle$
Just put a list of indices inside the ket, and read it as the tensor product of a list of basis vectors.
Bras work the same way — put anything inside them you want (all right, class…) as long as it specifies a vector. The difference is that the bra $\langle w\rvert$ denotes a vector in the dual space $V^*$. For example, given a basis for $V$, we may write $\langle i\rvert=\epsilon^i$ for a dual basis vector.
Putting a bra and a ket together means the same as evaluating the linear functional specified by the bra at the vector specified by the ket. Or we could remember that we can consider any vector in $V$ to be a linear functional on $V^*$, and read the bra-ket as an evaluation that way. The nice part about Dirac notation is that it doesn’t really privilege either viewpoint — both the bra and the ket enter on an equal footing.
Posted by John Armstrong | Algebra, Linear Algebra | 14 Comments
## Updates
As a much-anticipated visit approaches, my apartment is asymptotically approaching neatness. My former students’ comprehension of the word “asymptotically”, however, remains constant.
And this morning I received the proofs to a paper that was accepted for publication two years ago: “Functors Extending the Kauffman Bracket”. So yeah, I get out with it finally appearing in print, but in the meantime my career is shot.
Posted by John Armstrong | Uncategorized | 7 Comments
## Matrices and Forms II
Let’s pick up our discussion of matrices and forms and try to tie both views of our matrix together. We’re considering a bilinear form $B:V\otimes V\rightarrow\mathbb{F}$ on a vector space $V$ over the real or complex numbers, which we can also think of as a linear map from $V$ to its dual space $V^*$. We can also consider a sesquilinear form on a complex vector space, which is equivalent to an antilinear (conjugates scalars) map from the space to its dual.
We got a matrix by picking a basis $\left\{e_i\right\}$ for $V$ and plugging basis vectors into our form
$\displaystyle b_{ij}=B(e_i,e_j)$
And we also found that this is the matrix of the map from $V$ to $V^*$, written in terms of the basis $\left\{e_i\right\}$ and its dual basis $\left\{\epsilon^i\right\}$.
Now we have two bases of the same cardinality, so there is a (highly non-canonical) linear isomorphism from $V^*$ to $V$ which sends the basis vector $\epsilon^i$ to the basis vector $e_i$. If we compose this with the map from $V$ to $V^*$ given by the form $B$, we get a linear map from $V$ to itself, which we will also call $B$. If $B$ is sesquilinear, we use the unique antilinear isomorphism sending $\epsilon^i$ to $e_i$, and again get a linear map $B:V\rightarrow V$. The matrix of this linear map with respect to the basis $\left\{e_i\right\}$ is $b_i{}^j=b_{ij}$, just as before.
This seems sort of artificial, but there’s method here. Remember that we can come up with an inner product $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$ by simply declaring the basis $\left\{e_i\right\}$ to be orthonormal. Then the linear functional $\epsilon^i$ is given by $\langle e_i,\underline{\hphantom{X}}\rangle$. It’s this correspondence that is captured by both the non-canonical choice of inner product, and by the non-canonical choice of isomorphism.
But now we can extract the same matrix in a slightly different way. We start by hitting the basis vector $e_j$ with the new linear transformation $B$ to come up with a linear combination of the basis vectors. We then extract the appropriate component by using the inner product. In the end, we find
$\displaystyle\langle e_i,B(e_j)\rangle=b_{ij}=B(e_i,e_j)$
So in the presence of a given basis (or at least a given inner product) any bilinear (or sesquilinear) form corresponds to a linear transformation, and the matrix of the linear transformation with respect to the selected basis is exactly that of the form itself. On the other hand, if we have a linear transformation from $V$ to itself, we can stick it into the inner product as above and get a bilinear (or sesquilinear) form out.
Posted by John Armstrong | Algebra, Linear Algebra | 5 Comments
## Matrices and Forms I
Yesterday, we defined a Hermitian matrix to be the matrix-theoretic analogue of a self-adjoint transformation. So why should we separate out the two concepts? Well, it turns out that there are more things we can do with a matrix than represent a linear transformation. In fact, we can use matrices to represent forms, as follows.
Let’s start with either a bilinear or a sesquilinear form $B\left(\underline{\hphantom{X}},\underline{\hphantom{X}}\right)$ on the vector space $V$. Let’s also pick an arbitrary basis $\left\{e_i\right\}$ of $V$. I want to emphasize that this basis is arbitrary, since recently we’ve been accustomed to automatically picking orthonormal bases. But notice that I’m not assuming that our form is even an inner product to begin with.
Now we can define a matrix $b_{ij}=B(e_i,e_j)$. This completely specifies the form, by either bilinearity or sesquilinearity. And properties of such forms are reflected in their matrices.
For example, suppose that $H$ is a conjugate-symmetric sesquilinear form. That is, $H(v,w)=\overline{H(w,v)}$. Then we look at the matrix and find
$\displaystyle\begin{aligned}h_{ij}&=H\left(e_i,e_j\right)\\&=\overline{H\left(e_j,e_i\right)}\\&=\overline{h_{ji}}\end{aligned}$
so $H$ is a Hermitian matrix!
Now the secret here is that the matrix of a form secretly is the matrix of a linear transformation. It’s the transformation that takes us from $V$ to $V^*$ by acting on one slot of the form, and written in terms of the basis $e_i$ and its dual. Let me be a little more explicit.
When we feed a basis vector into our form $B$, we get a linear functional $B(e_i,\underline{\hphantom{X}})$. We want to write that out in terms of the dual basis $\left\{\epsilon^j\right\}$ as a linear combination
$\displaystyle B(e_i,\underline{\hphantom{X}})=b_{ik}\epsilon^k$
So how do we read off these coefficients? Stick another basis vector into the form!
$\displaystyle\begin{aligned}B(e_i,e_j)&=b_{ik}\epsilon^k(e_j)\\&=b_{ik}\delta^k_j\\&=b_{ij}\end{aligned}$
which is just the same matrix as we found before.
Posted by John Armstrong | Algebra, Linear Algebra | 12 Comments
## Self-Adjoint Transformations
Let’s now consider a single inner-product space $V$ and a linear transformation $T:V\rightarrow V$. Its adjoint is another linear transformation $T^*:V\rightarrow V$. This opens up the possibility that $T^*$ might be the same transformation as $T$. If this happens, we say that $T$ is “self-adjoint”. It then satisfies the adjoint relation
$\displaystyle\langle v,T(w)\rangle=\langle T(v),w\rangle$
What does this look like in terms of matrices? Since we only have one vector space we only need to pick one orthonormal basis $\left\{e_i\right\}$. Then we get a matrix
$\displaystyle\begin{aligned}t_i^j&=\langle e_j,T(e_i)\rangle\\&=\langle T(e_j),e_i\rangle\\&=\overline{\langle e_i,T(e_j)\rangle}\\&=\overline{t_j^i}\end{aligned}$
That is, the matrix of a self-adjoint transformation is its own conjugate transpose. We have a special name for this sort of matrix — “Hermitian” — even though it’s exactly equivalent to self-adjointness as a linear transformation. If we’re just working over a real vector space we don’t have to bother with conjugation. In that case we just say that the matrix is symmetric.
Over a one-dimensional complex vector space, the matrix of a linear transformation $T$ is simply a single complex number $t$. If $T$ is to be self-adjoint, we must have $t=\bar{t}$, and so $t$ must be a real number. In this sense, the operation of taking the conjugate transpose of a complex matrix (or the simple transpose of a real matrix) extends the idea of conjugating a complex number. Self-adjoint matrices, then, are analogous to real numbers.
Posted by John Armstrong | Algebra, Linear Algebra | 9 Comments
## The Matrix of the Adjoint
I hear joints popping as I stretch and try to get back into the main line of my posts.
We left off defining what we mean by a matrix element of a linear transformation. Let’s see how this relates to adjoints.
We start with a linear transformation $T:V\rightarrow W$ between two inner product spaces. Given any vectors $v\in V$ and $w\in W$ we have the matrix element $\langle w,T(v)\rangle_W$, using the inner product on $W$. We can also write down the adjoint transformation $T^*:W\rightarrow V$, and its matrix element $\langle v,T^*(w)\rangle_V$, using the inner product on $V$.
But the inner product on $W$ is (conjugate) linear. That is, we know that the matrix element $\langle w,T(v)\rangle_W$ can also be written as $\overline{\langle T(v),w\rangle_W}$. And we also have the adjoint relation $\langle v,T^*(w)\rangle_V=\langle T(v),w\rangle_W$. Putting these together, we find
$\displaystyle\begin{aligned}\langle v,T^*(w)\rangle_V&=\langle T(v),w\rangle_W\\&=\overline{\langle w,T(v)\rangle_W}\end{aligned}$
So the matrix elements of $T$ and $T^*$ are pretty closely related.
What if we pick whole orthonormal bases $\left\{e_i\right\}$ of $V$ and $\left\{f_j\right\}$ of $W$? Now we can write out an entire matrix of $T$ as $t_i^j=\langle f_j,T(e_i)\rangle_W$. Similarly, we can write a matrix of $T^*$ as
$\displaystyle\begin{aligned}\left(t^*\right)_j^i&=\langle e_i,T^*(f_j)\rangle_V\\&=\overline{\langle f_j,T(e_i)\rangle_W}\\&=\overline{t_i^j}\end{aligned}$
That is, we get the matrix for the adjoint transformation by taking the original matrix, swapping the two indices, and taking the complex conjugate of each entry. This “conjugate transpose” operation on matrices reflects adjunction on transformations.
Posted by John Armstrong | Algebra, Linear Algebra | 2 Comments
## Hiatus
This week’s been hectic, what with interviews (which I can’t talk about but I think went okay) and driving back to Kentucky. And now I’m about to spend two weeks up at Mammoth Cave. I should have some internet access, so I’ll try to keep up with lower-content features, but real mathy stuff is off the table for a while. You can always follow my twitter stream to keep tabs, and find links to pictures I take underground, assuming that the wireless at Hamilton Valley really works. But of course, the best laid plans of mice and men…
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathoverflow.net/questions/61252/why-do-bernoulli-numbers-arise-everywhere/61278
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## Why do Bernoulli numbers arise everywhere?
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I have seen Bernoulli numbers many times, and sometimes very surprisingly. They appear in my textbook on complex analysis, in algebraic topology, and of course, number theory. Things like the criteria for regular primes, or their appearance in the Todd class, zeta value at even numbers looks really mysterious for me. (I remember in Milnor's notes about characteristic class there is something on homotopy group that has to do with Bernoulli numbers, too, but I don't recall precisely what that is. I think they also arise in higher K-theory.)
The list can go on forever. And the wikipedia page of Bernoulli number is already quite long.
My question is, why do they arise everywhere? Are they a natural thing to consider?
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p.s.----(maybe this should be asked in a separate question)
Also, I've been wondering why it is defined as the taylor coefficient of the particular function x/(e^x-1), was this function important? e.g. I could have taken the coefficient of the series that defines the L-genus, namely $\dfrac{\sqrt{z}}{\text{tanh}\sqrt{z}}$, which only amounts to change the Bernoulli numbers by some powers of 2 and some factorial. I guess many similar functions will give you the Bernoulli numbers up to some factor. Why it happen to be the function x/(e^x-1)?
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According to history of mathematics, at the beginning of Bernoulli's original idea, he did not begin defined as the Taylor coefficients of the function x/(e^x-1) at 0. – yaoxiao Apr 11 2011 at 2:56
That's a good point. So do you know how did he define it and what's his motivation? – 36min Apr 11 2011 at 4:19
Just a minor remark: the space of the areas you have mentioned might be simply connected. For example, the criteria for regular primes, the zeta values at even integers, and higher K-theory - these are all closely related areas. If you are willing to take on board everything we know and everything we conjecture to be true, then it follows that Bernoulli numbers appear in one of these if and only if they appear in all of them. Somewhat related is my answer here: mathoverflow.net/questions/45376/… – Alex Bartel Apr 11 2011 at 4:42
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@Alex, I take it that by "simply connected" you mean "connected in a simple way". – gowers Apr 11 2011 at 8:32
@Tim You are right, I guess I didn't mean simply connected, but rather a complete graph or something like that. – Alex Bartel Apr 11 2011 at 12:54
show 1 more comment
## 6 Answers
I don't know of a universal theory of all places where Bernoulli numbers arise, but Euler-Maclaurin summation explains many of their more down-to-earth occurrences.
The heuristic explanation (due to Lagrange) is as follows. The first difference operator defined by $\Delta f(n) = f(n+1)-f(n)$ and summation are inverses, in the same sense in which differentiation and integration are inverses. This just amounts to a telescoping series: $\sum_{a \le i < b} \Delta f(i) = f(b) - f(a)$.
Now by Taylor's theorem, $f(n+1) = \sum_{k \ge 0} f^{(k)}(n)/k!$ (under suitable hypotheses, of course). If we let $D$ denote the differentation operator defined by $Df = f'$, and $S$ denote the shift operator defined by $Sf(n) = f(n+1)$, then Taylor's theorem tells us that $S = e^D$. Thus, because $\Delta = S-1$, we have $\Delta = e^D - 1$.
Now summing amounts to inverting $\Delta$, or equivalently applying $(e^D-1)^{-1}$. If we expand this in terms of powers of $D$, the coefficients are Bernoulli numbers (divided by factorials). Because of the singularity at "$D=0$", the initial term involves antidifferentiation $D^{-1}$, i.e., integration. Thus, we have expanded a sum as an integral plus correction terms involving higher derivatives, with Bernoulli number coefficients.
Specifically, $$\sum_{a \le i < b} f(i) = \int_a^b f(x) \, dx + \sum_{k \ge 1} \frac{B_k}{k!} (f^{(k-1)}(b) - f^{(k-1)}(a)).$$ (Subtracting the values at $b$ and $a$ just amounts to the analogue of turning an indefinite integral into a definite integral.)
This equation isn't literally true in general: the infinite sum usually won't converge and there's a missing error term. However, it is true when $f$ is a polynomial, and one can bootstrap from this case to the general one using the Peano kernel trick.
So from this perspective, the reason why $t/(e^t-1)$ is a natural generating function to consider is that we sometimes want to invert $e^t-1$ (the factor of $t$ is just to make it holomorphic), and the most important reason I know of to invert it is that we want to invert $\Delta = e^D-1$.
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If I recall correctly, the infinite sum converges to the correct answer also for polynomials times exponentials. – Allen Knutson Apr 11 2011 at 3:42
You can do this in higher dimensions too, integrating over polytopes. – Steve Huntsman Apr 11 2011 at 12:44
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See also this question: mathoverflow.net/questions/10667 – aorq Apr 12 2011 at 0:34
Also umbrally $p(B(x)+1)-p(B(x))={p}'(x)$ where $B(x)$ are the Bernoulli polynomials and $p(x)$ any polynomial, so you might expect them to pop up in approximation and linearization problems. – Tom Copeland Sep 18 at 2:56
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The main reason I know for the appearance of Bernoulli numbers is the one Henry Cohn already explained: we'd like to invert the difference operator $e^D - 1$, so we'd like to expand $1/(e^D - 1)$ as a Taylor series. But $1/(e^x - 1)$ doesn't have a Taylor series, because it has a pole at the origin. It has a perfectly nice Laurent series, but just to make things more obscure people prefer to discuss the Taylor series of $x/(e^x - 1)$. And the coefficients of this are called Bernoulli numbers.
I understand how Bernoulli numbers are used to compute $\sum_{i=1}^n i^k$ and how they show up in formulas for the Riemann zeta function.
However, Alain Connes loses me here:
• Alain Connes, Andre Lichnerowicz and Marcel Paul Schutzenberger, A Triangle of Thoughts, AMS, Providence, 2000.
He points out that if $H$ is the Hamiltonian for some sort of particle in a box and $\beta$ is the inverse temperature,
$$1/(1 - e^{-\beta H}) = 1 + e^{-\beta H} + e^{-2 \beta H} + \cdots$$
is the operator you take the trace of to get the partition function for a collection of an arbitrary number of particles of this sort. And he claims that pondering this explains all the appearances of $x/(1 - e^x)$ and the Bernoulli numbers in topology!
Does anyone understand that? I imagine he's hinting at some relation between characteristic classes, the heat equation, the Laplacian on differential forms, and things like that. But I've never understood how these pieces are supposed to fit together.
And here's something that remains more mysterious to me. The paper by Kervaire and Milnor has a cool formula for the order of the group of smooth structures on the (4n-1)-sphere for n > 1. It's:
$$2^{2n-4} (2^{2n-1} - 1) P(4n-1) B(n) a(n) / n$$
where:
$P(k)$ is the order of the $k$th stable homotopy group of spheres
$B(k)$ is the $k$th Bernoulli number, in the sequence 1/6, 1/30, 1/42, 1/30, 5/66, 691/2730, 7/6, ...
$a(k)$ is 1 or 2 according to whether k is even or odd
How do the Bernoulli numbers weasel their way into this game?
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I think I have a rough idea of how Bernoulli numbers get into exotic spheres. There's this "J-homomoprhism" Z x BO --> Pic(S), which sends a (stable) real vector bundle to the (stable) spherical fibration given by one-point compactifying the fibers. The Spivak normal fibration over S^n gives a canonical map S^n --> Pic(S), and finding manifold structures on S^n is tied to lifting this map along J (so saying that the normal fibration actually came from a normal vector bundle). This lifting problem comes down to understanding what J does on homotopy groups. This in turn can be understood ... – Dustin Clausen Apr 11 2011 at 16:19
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in terms of Pontryagn-Thom theory: a class in the kernel of J on pi_n is represented by an "exotic" null-bordism of S^n. These correspond to almost parallelizable manifolds (manifolds parallelizable away from a fixed point), and once you've got one of those most of your characteristic classes vanish, and (for instance) the integrality of the A-hat genus will tell you something about just one particular Bernoulli number. Well OK, but what does this have to do with Euler-Maclaurin summation?? :) – Dustin Clausen Apr 11 2011 at 16:23
John discussed the relevant passage of 'A Triangle of Thoughts' at greater length here golem.ph.utexas.edu/category/2008/02/… – David Corfield Apr 12 2011 at 8:27
Also Harer and Zagier have shown that the orbifold Euler characteristic of the moduli space $\mathcal M_{g,n}$ is $(-1)^{n-1}\frac{(2g+n-3)!}{(2g-2)!} \zeta(1-2g)$. – Tom Copeland Sep 18 at 2:46
In algebraic topology one key point is as follows. The complex $K$-theory spectrum has homotopy groups $KU_{\ast}=\mathbb{Z}[u,u^{-1}]$, with $u$ in degree two. This ring maps in an obvious way to $\pi_{\ast}(H\wedge KU)$, and it is not hard to calculate that the resulting map $\mathbb{Z}[u,u^{-1}]\to\pi_{\ast}(H\wedge KU)$ induces an isomorphism $\mathbb{Q}[u,u^{-1}]\to\pi_{\ast}(H\wedge KU)$. The ring $(H\wedge KU)^0(\mathbb{C}P^\infty)$ can be described as $\mathbb{Q}[[ux]]$ or as $\mathbb{Q}[[y]]$, where $x$ comes from $H^2(\mathbb{C}P^\infty)$ and $y$ comes from $KU^0(\mathbb{C}P^\infty)$. Specifically, if we let $L$ denote the tautological line bundle, then $y$ can be taken to be the $K$-theory class of the virtual bundle $L-1$. It then works out that $y=e^x-1$, so $x/y$ is the Bernoulli series. The Bernoulli numbers occur as coefficients of $x^k/k!$ rather than $x^k$ itself, which suggests that one should work with $\Omega S^3$ rather then $\mathbb{C}P^\infty$: there is a canonical map $\Omega S^3\to\mathbb{C}P^\infty$ using which we can identify $H^*(\Omega S^3)$ with the ring of all series of the form $\sum_ka_kx^k/k!$ with $a_k\in\mathbb{Z}$. All this is of course linked with Adams's treatment of the $J$-homomorphism and the $e$-invariant. However, I think that much of this is still mysterious, at least to me. I think there are some missing ingredients involving the relationship between $R$ and $gl_1(R)$ for various $E_\infty$ ring spectra $R$, particularly those related to surgery and the $J$-homomorphism. There is a lot of literature about this kind of thing from the 1970s but I have not managed to extract the answers that I wanted.
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I remember this question coming up one tea time when I was sitting next to Frank Adams, and he said the reasons for the occurrence of Bernoulli numbers in topology were "trivial". Makes me wish I'd asked "compared to what?" – Charles Matthews Apr 11 2011 at 14:44
I have been told (by J.H. Conway) that Bernoulli numbers were first discovered by Faulhauber. See the http://en.wikipedia.org/wiki/Faulhaber's_formula wikipedia article for details. This reference hardly answers your question, but one possible characterization is that they are useful in summing nth powers. That fact alone indicates that their ubiquity is quite natural to expect.
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Another way where they show up is in Lie theory. If you want to compute the derivative of the exponential map (of a Lie group) you encounter the function $x/(e^x - 1)$ quite inevitably. The already posted questions can partly be viewed as incarnations of this. THis results also in the appearence of the Bernoulli numbers in the BCH series, which is of course of fundamental importance far beyond the usage in Lie algebra theory...
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One of ways for understanding the power of Bernoulli numbers and your question is that Bernoulli numbers grow faster than any geometric series.see http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2023%20Bernoulli%20numbers.pdf
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http://www.physicsforums.com/showthread.php?s=4b7c6c19868f87b41f69b69f91f3ad9a&p=3986308
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Physics Forums
## Which method of integration for sin^2(x) dx
During maths class last semester this integral came up in the course of discussion and my lecturer gave a quick outline of how to solve it but I didn't grasp it at the time and we moved on. I'd like to know how to do it though! The integral is:
$\int sin^{2}(x)$
The next step was:
$\int sin(x).(-cos(x)')$
where $-cos(x)'$ denotes the derivative of $-cos(x)$ and the implication was that $sin^{2}(x)$ is the result of applying the chain rule to whatever compound function it is the derivative of. I can't remember the result but further steps were skipped and he jumped straight to the answer at this point.
I'm not sure how to proceed here, because the chain rule implies that $\frac{dy}{dx} f(g(x)) = f'(g(x) . g'(x)$, and in my example $g(x) = x$ and $g'(x) = 1$ which doesn't help me end up with $sin(x)$ as required.
Could anyone poke me in the right direction to solve this? =)
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I'd forget the chain rule and try to re-write $\sin^2x$ as something I could integrate, if I were you. It might be possible by parts, too, but I haven't actually tried that so I wouldn't swear to it.
The method I was taught was to use the double angle formula for cosine. I really have no idea what your teacher is trying to do.
Mentor
## Which method of integration for sin^2(x) dx
Quote by Adyssa I'm not sure how to proceed here, because the chain rule implies that $\frac{dy}{dx} f(g(x)) = f'(g(x) . g'(x)$, and in my example $g(x) = x$ and $g'(x) = 1$ which doesn't help me end up with $sin(x)$ as required. Could anyone poke me in the right direction to solve this? =)
You don't want the chain rule. You want the product rule, the integration equivalent of which is integration by parts.
Before nudging you in the right direction, there is a much easier way to find $\int sin^2x\,dx$. Use the identity $\sin^2x = (\frac{1-\cos(2x)} 2$. The right hand side is easy to integrate.
The product rule is $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$. Transforming to integration,$$\int (f(x)g(x))' dx = f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx$$
Note that this is integration by parts, just written a bit differently. You are using $f(x)=\sin x$ and $g(x)=-\cos x$. Plug these in. You'll eventually need a different trig identity, $\sin^2x + \cos^2x = 1$, to finish it off.
Thanks for the replies. I'm really awful at remembering trig identities (and remembering in general =S), I had a large gap between high school and university maths, so I have to re-learn a lot of the basics as I go along. I solved it by using the double angle identity $sin^{2}x = \frac{1 - cos(2x)}{2}$ and integration by substitution - $u = 2x, du = 2dx, dx = \frac{1}{2}du$, and my result was $\frac{x}{2} - \frac{1}{4}sin(2x)$. I took the derivative to check my answer and it worked, so that's great! I'm going to have a go at the integration by parts method now.
Mentor
Adyssa, I had a typo (matho) in my previous post.
Quote by Adyssa I took the derivative to check my answer and it worked, so that's great!
It is always a good idea to double check your work. Good job!
You can try another technique... let I=∫sin$^{2}$xdx ---------(i) or I=∫dx-∫cos$^{2}$xdx or I=x-∫cos$^{2}$xdx +c$_{1}$ -------(ii) Adding (i) and (ii), 2I=x-∫[cos$^{2}$x-sin$^{2}$x]dx +c$_{1}$ 2I=x-∫cos2xdx +c$_{1}$ 2I=x-1/2 sin2x +c I=x/2 - 1/4 sin2x+c
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http://mathhelpforum.com/calculus/112525-increasing-decreasing-functions.html
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# Thread:
1. ## increasing and decreasing functions
I am to determine the interval on which f(x)=x-x^-2 is increasing. I came up with the derivative of f`(x) =x-x^-2 as f`(x)=1+2/x^3. Having said that, the graph on the left side of the y axis seems to travel in a positive direction until it gets to the y axis, then goes to -infinity. The graph on the right side of the y axis seems to go from o to infinity. Am I doing this correctly? Is the increasing interval (o, to infinity)?
2. Originally Posted by bosmith
I am to determine the interval on which f(x)=x-x^-2is increasing. I came up with the derivative of f`(x) =x-x^-2
Poorly written! You do not mean to say that f'(x) is $x-x^{-2}$ and that is what "=" means! You mean $f(x)= x-x^{-2}$
as f`(x)=1+2/x^3. Having said that, the graph on the left side of the y axis seems to travel in a positive direction until it gets to the y axis, then goes to -infinity. The graph on the right side of the y axis seems to go from o to infinity. Am I doing this correctly? Is the increasing interval (o, to infinity)?
Well, you've already done the work. The function is increasing as long as $f'(x)= x- x^{-2}> 0$. Since $x^2> 0$ for all real (non-zero) x, we can multiply on both sides by $x^2$ and get [tex]x^3- x> 0[\math]. That factors as $x(x^2- 1)= x(x-1)(x+1)> 0$. The simplest way to determine where the derivative is positive or negative is to recognize that it can change from one to the other only where it is negative. And that occurs only at x=-1, x= 0, and x= 1. Choose one value for x in each of the intervals $-\infty< x< -1$, -1< x< 0, 0< x< 1, and $1< x< \infty$ to see whether the derivative is positive or negative in that interval.
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http://quant.stackexchange.com/questions/2152/how-to-perform-basic-integrations-with-the-ito-integral/2153
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# How to perform basic integrations with the Ito integral?
From the text book Quantitative Finance for Physicists: An Introduction (Academic Press Advanced Finance) I have this excercise:
Prove that
$$\int_{t_1}^{t_2}W(s)^ndW(s)=\frac{1}{n+1}[W(t_2)^{n+1}-W(t_1)^{n+1}]-\frac{n}{2}\int_{t_1}^{t_2}W(s)^{n-1}ds$$
Hint: Calculate $d(W^{n+1})$ using Ito's lemma.
This is my calculation:
I use Ito's Lemma and use, as the text book does, the simplified case $\mu=0$, $\sigma=1$. So Ito's lemma reduces to:
$$dF=dt+2WdW$$
Now I use the Ito Lemma here like this:
$$\int_{t_1}^{t_2}W(s)^ndW(s)=\int_{t_1}^{t_2}W(s)^{n-1}W(s)dW(s)$$
Because $\mu=0$, $\sigma=1$, we have $F=W^2$ and therefore the integral equals: $$\int_{t_1}^{t_2}F^\frac{n-1}{2}\frac{dF}{2}-\int_{t_1}^{t_2}F^\frac{n-1}{2}\frac{dt}{2}$$ Further simplification: $$\frac{1}{2}\frac{2}{n+1}F^\frac{n+1}{2}]_{t_1}^{t_2}-\frac{1}{2}W^{n-1}(t)dt=\frac{1}{n+1}W^{n+1}]_{t_1}^{t_2}-\frac{1}{2}W(t)^{n-1}(t)dt$$ Putting everything together obviously yields: $$\int_{t_1}^{t_2}W(s)^ndW(s)=\frac{1}{n+1}[W(t_2)^{n+1}-W(t_1)^{n+1}]-\frac{1}{2}\int_{t_1}^{t_2}W(s)^{n-1}ds$$
The second term lacks a factor $n$.
What am I doing wrong? Or is the book wrong? (By the way for $n=1$ this is consistent with the book. For $n=1$ I was also able to calculate the integral using summation $\lim_n\sum$.... But this seems too complicated for general case.)
Besides I have problems understanding why I have to use the differential equation. Or do I have to see $W$ as $W=W(F,t)$?
Edit: Removed some lines...
-
your assumptions about $\mu$ and $\sigma$ are unclear to me. but see my answer – SRKX♦ Oct 12 '11 at 20:00
The text book does in that chapter most calculations with this simplified case. – Philip Oct 12 '11 at 20:07
also, for latter readers, you might want to tell us what textbook you're using. – SRKX♦ Oct 12 '11 at 20:11
the formula you use for ito is the particular case $f(x)=x^2$ – SRKX♦ Oct 12 '11 at 20:13
I added the reference. I see, in fact the book isn't really clear about Ito. It starts by deriving it using the Taylor expansion of $df$ and then continues simplifying it step by step. – Philip Oct 12 '11 at 20:25
## 1 Answer
I think you should see the hint as follows:
$$d(W_t^{n+1})=d(f(W_t))$$ with $$f(x)=x^{n+1}$$
Apply Ito:
$$d(W_t^{n+1}) = f'(W_t)dW_t + \frac{1}{2} f''(W_t) d<W>_t$$
$$d(W_t^{n+1}) = (n+1) W_t^n dW_t + \frac{1}{2} n (n+1) W_t^{n-1} dt$$
If you integrate, you get:
$$W_{t_2}^{n+1}-W_{t_1}^{n+1}=(n+1) \int_{t_1}^{t_2} W_t^n dW_t+ \frac{1}{2} n (n+1) \int_{t_1}^{t_2} W_t^{n-1} dt$$
You divide both sides by $(n+1)$ and you're done.
I'm not sure how you use your Ito's lemma, maybe you've been fooled by some simplified version of some books. I'd recommend Elementary Stochastic Calculus with Finance in View if you want a good introduction to the field.
As for you assumptions, $\mu=0$ and $\sigma=1$ are the properties of a Brownian motion after just a single step $W_1$, by definition.
Here the purpose of the exercise is in my opinion to show you that by applying Ito's lemma, you might end up finding interesting properties (one of the most famous one was shown in this post).
-
Thanks a lot! Though I still have to find my error... – Philip Oct 12 '11 at 20:05
@Philip: just added more details to my answer because honestly I'm not sure what you're doing exactly in your way to tackle the problem. – SRKX♦ Oct 12 '11 at 20:10
@Philip: found your mistake, see the last comment on the question. – SRKX♦ Oct 12 '11 at 20:20
Great, thanks again, also for the references. – Philip Oct 12 '11 at 20:26
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http://johncarlosbaez.wordpress.com/2011/03/04/network-theory-part-1/
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# Azimuth
## Network Theory (Part 1)
As a mathematician who has gotten interested in the problems facing our planet, I’ve been trying to cook up some new projects to work on. Over the decades I’ve spent a lot of time studying quantum field theory, quantum gravity, n-categories, and numerous pretty topics in pure math. My accumulated knowledge doesn’t seem terribly relevant to my new goals. But I don’t feel like doing a complete ‘brain dump’ and starting from scratch. And my day job still requires that I prove theorems.
#### Green Mathematics
I wish there were a branch of mathematics—in my dreams I call it green mathematics—that would interact with biology and ecology just as fruitfully as traditional mathematics interacts with physics. If the 20th century was the century of physics, while the 21st is the century of biology, shouldn’t mathematics change too? As we struggle to understand and improve humanity’s interaction with the biosphere, shouldn’t mathematicians have some role to play?
Of course, it’s possible that when you study truly complex systems—from a living cell to the Earth as a whole—mathematics loses the unreasonable effectiveness it so famously has when it comes to simple things like elementary particles. So, maybe there is no ‘green mathematics’.
Or maybe ‘green mathematics’ can only be born after we realize it needs to be fundamentally different than traditional mathematics. For starters, it may require massive use of computers, instead of the paper-and-pencil methods that work so well in traditional math. Simulations might become more important than proofs. That’s okay with me. Mathematicians like things to be elegant—but one can still have elegant definitions and elegant models, even if one needs computer simulations to see how the models behave.
Perhaps ‘green mathematics’ will require a radical shift of viewpoint that we can barely begin to imagine.
It’s also possible that ‘green mathematics’ already exists in preliminary form, scattered throughout many different fields: mathematical biology, quantitative ecology, bioinformatics, artificial life studies, and so on. Maybe we just need more mathematicians to learn these fields and seek to synthesize them.
I’m not sure what I think about this ‘green mathematics’ idea. But I think I’m getting a vague feel for it. This may sound corny, but I feel it should be about structures that are more like this:
than this:
I’ve spent a long time exploring the crystalline beauty of traditional mathematics, but now I’m feeling an urge to study something slightly more earthy.
#### Network Theory
When dreaming of grand syntheses, it’s easy to get bogged down in vague generalities. Let’s start with something smaller and more manageable.
Network theory, and the use of diagrams, have emerged independently in many fields of science. In particle physics we have Feynman diagrams:
In the humbler but more practical field of electronics we have circuit diagrams:
Throughout engineering we also have various other styles of diagram, such as bond graphs:
I’ve already told you about Petri nets, which are popular in computer science… but also nice for describing chemical reactions:
‘Chemical reaction networks’ do a similar job, in a more primitive way:
Chemistry shades imperceptibly into biology, and biology uses so many styles of diagram that an organization has tried to standardize them:
• Systems Biology Graphical Notation (SBGN) homepage.
SBGN is made up of 3 different languages, representing different visions of biological systems. Each language involves a comprehensive set of symbols with precise semantics, together with detailed syntactic rules how maps are to be interpreted:
1) The Process Description language shows the temporal course of biochemical interactions in a network.
2) The Entity Relationship language lets you to see all the relationships in which a given entity participates, regardless of the temporal aspects.
3) The Activity Flow language depicts the flow of information between biochemical entities in a network.
Biology shades into ecology, and in the 1950s, Howard T. Odum developed the ‘Energy Systems Language’ while studying tropical forests. Odum is now considered to be the founder of ‘systems ecology’. If you can get ahold of this big fat book, you’ll see it’s packed with interesting diagrams describing the flow of energy through ecosystems:
• Howard T. Odum, Systems Ecology: an Introduction, Wiley-Interscience, New York, 1983.
His language is sometimes called ‘Energese’, for short:
The list goes on and on, and I won’t try for completeness… but we shouldn’t skip probability theory, statistics and machine learning! A Bayesian network, also known as a “belief network”, is a way to represent knowledge about some domain: it consists of a graph where the nodes are labelled by random variables and the edges represent probabilistic dependencies between these random variables. Various styles of diagrams have been used for these:
And don’t forget neural networks!
#### What Mathematicians Can Do
It’s clear that people from different subjects are reinventing the same kinds of diagrams. It’s also clear that diagrams are being used in a number of fundamentally different ways. So, there’s a lot to sort out.
I already mentioned one attempt to straighten things out: Systems Biology Graphical Notation. But that’s not the only one. For example, in 2001 the International Council on Systems Engineering set up a committee to customize their existing Unified Modeling Language and create something called Systems Modeling Language. This features nine types of diagrams!
So, people are already trying to systematize the use of diagrams. But mathematicians should join the fray.
Why? Because mathematicians are especially good at soaring above the particulars and seeing general patterns. Also, they know ways to think of diagrams, not just as handy tools, but as rigorously defined structures that you can prove theorems about… with the help of category theory.
I’ve written a bit about diagrams already, but not their ‘green’ applications. Instead, I focused on their applications to traditional subjects like topology, physics, logic and computation:
• John Baez and Aaron Lauda, A prehistory of n-categorical physics, to appear in Deep Beauty: Mathematical Innovation and the Search for an Underlying Intelligibility of the Quantum World, ed. Hans Halvorson, Cambridge U. Press.
• John Baez and Mike Stay, A Rosetta stone: topology, physics, logic and computation, in New Structures for Physics, ed. Bob Coecke, Lecture Notes in Physics vol. 813, Springer, Berlin, 2011, pp. 95-174.
It would be good to expand this circle of ideas to include chemistry, biology, ecology, statistics, and so on. There should be a mathematical theory underlying the use of networks in all these disciplines.
I’ve started a project on this with Jacob Biamonte, who works on two other applications of diagrams, namely to quantum computation and condensed matter physics:
• Jacob D. Biamonte, Stephen R. Clark and Dieter Jaksch, Categorical tensor networks.
So far we’ve focused on one aspect: stochastic Petri nets, which are used to describe chemical reactions and also certain predator-prey models in quantitative ecology. In the posts to come, I want to show how ideas from quantum field theory be used in studying stochastic Petri nets, and how this relates to the ‘categorification’ of Feynman diagram theory.
This entry was posted on Friday, March 4th, 2011 at 7:33 am and is filed under biology, mathematics, networks, physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### 57 Responses to Network Theory (Part 1)
1. Tim van Beek says:
…it may require massive use of computers, instead of the paper-and-pencil methods that work so well in traditional math. Simulations might become more important than proofs.
I’ve seen thousands of mistakes made because people accepted a wrong result of a computer or misinterpreted the results, although they could have spotted the mistake by analyzing the problem with pen and paper.
It’s like a class in complex analysis where the students learn how to evaluate integrals of real functions over the real axis using residue calculus, sooner or later someone will make a mistake (or use a computer algebra program that has a bug) and get a complex number as a result of a real integral of a real function. Or a finite number for a divergent integral. It will always be useful to be able to tell that such a result has to be wrong, based on traditional pen-and-paper techniques :-)
(To me computers are nothing more than advanced pen-and-papers anyway.)
For example, in 2001 the International Council on Systems Engineering set up a committee to customize their existing Unified Modeling Language and create something called Systems Modeling Language. This features nine types of diagrams!
One of the main problems is that these diagrams – when used for software engineering – are supposed to model interactions, processes and dependencies of all kinds, from user-software interactions that are very hard to formalize, to collaboration diagrams of objects in an object-oriented system, where the very meaning of “collaboration” takes a special connotation for every line that is drawn. One big challenge in software engineering is to figure out what aspects of a complex system can be formalized and to what level, and what can not.
The far reaching goal of the UML was to enable software engineers to generate a large part of the code of a new software system from UML diagrams, there exist a lot of tools today that can generate DDL scripts (DDL = database description language, specifies the table layout of relational databases) from entity-relationship diagrams, and code from class diagrams, and vice versa. But usually these are only intermediate steps of the development, and a lot of work has to be done afterwards, because the UML models are usually much too coarse.
• WebHubTel says:
I would agree with Tim’s sentiment.
The problem with UML is that, first and foremost, it is a tool designed to make money for the software tool vendors. So the vendors make sure that it is easy to use and it allows the users of the tool to take shortcuts and be loose with the diagramming rules. (Hey, it’s only a diagram, right?) But, because it is so loose with obeying the rules, especially WRT the activity diagrams and sequence diagrams, the end result is a diagram that is ambiguous and in many cases useless.
For example, Activity Diagrams include Petri Nets as a diagramming capability, yet it does not include any of the formal semantics of Petri Nets. So the result is that the software designers enjoy playing with the tool, don’t have to think too hard, and the vendor sells more of the tools as the word spreads.
Yet, the software developers still have to debug their code. The upshot is that many people consider that UML does not buy a development team that much, but it has that marketing muscle and high adoption levels behind it.
The alternative is to use something much more formal, but no one wants to buy the stuff, because then they will actually need to hire some methodical and thorough designers. Since there are not enough of these people the tool does not get enough sales and disappears from the market.
That is why we have UML; it makes the most people happy.
• Giampiero Campa says:
I agree with both of you.
I also think that one of the limitations of UML seems to be that it is not natural to model systems governed by differential equations in that formalism, and therefore it’s hard to model/simulate a system that it’s not just software but includes nontrivial physical subsystems.
This is one of the reasons why both the Automotive and Aerospace industries have long adopted Model Based Design as a development standard.
This seems to work better also for generating the code, i think for the simple reason that it forces the designer to gradually refine things (while writing some code) early on in the process, in order to simulate the whole system.
Now, while general purpose software tend to interface less with physical systems, it does interface much more with humans, which is a problem because you don’t have a mathematical model of the user, so you can’t simulate how your software behaves before deploying it. So cycling through a lot of beta versions is basically your only answer in that case, while i still think that formalisms such as UML do help a little.
• Siggy says:
As to use of computers over pen and paper… Yes, computers bring with them a laziness that leads to mistakes. But with many nonlinear and stochastic systems there is no other option whatsoever. Complex analysis is one thing. Complex systems are quite another beast, for which the tricks of traditional P.D.E. are often wholly impotent.
2. streamfortyseven says:
It’s a field of inquiry with a long history starting perhaps with Leibniz: http://en.wikipedia.org/wiki/Characteristica_universalis
Perhaps one way to begin to analyze the problem could be to use a “pattern language” (see http://en.wikipedia.org/wiki/Pattern_language) to look at wholes and the parts which comprise them, and the subparts which make up the parts, and continue on down to the atomic level of parts which can no longer be subidivided.
Here’s some articles on this topic:
Anatomy of a Pattern Language
http://www.designmatrix.com/pl/anatomy.html
Pattern Language as applied to architecture and urban design:
http://www.patternlanguage.com/leveltwo/patternsframegreen.htm?/leveltwo/../apl/twopanelnlb.htm
Pattern Language as applied to software engineering:
parlab.eecs.berkeley.edu/wiki/_media/patterns/paraplop_g1_1.pdf
As applied to permaculture design:
http://www.holocene.net/dissertation.htm
As applied to forest garden design:
http://appleseedpermaculture.com/forest-gardening-vision-pattern-language/
• WebHubTel says:
Patterns => also known as “design rules” in engineering circles.
http://en.wikipedia.org/wiki/Design_rule_checking
• streamfortyseven says:
As applied to musical improvisation:
http://doc.gold.ac.uk/~ma503am/essays/petrol-draft.pdf
and
http://smcnetwork.org/files/proceedings/2010/39.pdf
and a platform on which to create a pattern language for music and other things:
http://en.flossmanuals.net/PureData/Introduction
as applied to design of ecosystems:
http://www.designmatrix.com/pl/ecopl/index.html
and
http://www.irl.ethz.ch/plus/people/agrtrega/wissen_2010
a pattern language for querying directed acyclic graphs with several distinctive and innovative characteristics:
arantxa.ii.uam.es/~ssantini/work/papers/recent/s613_jisbd_grafos.pdf
and another platform (mac only, so far):
http://impromptu.moso.com.au/
• streamfortyseven says:
As applied to design of peer-to-peer overlay networks:
http://sct.ethz.ch/projects/student_docs/Dominik_Grolimund/Dominik_Grolimund_SA_paper.pdf
• John Baez says:
Thanks for all these links, streamfortyseven. Ever since my high school pal John Garrahan, who became an architect, gave me a copy of this book:
• Christopher Alexander, A Pattern Language.
I’ve been fond of this approach to architecture. My mother is a huge fan of Frank Lloyd Wright, and built a house in roughly that style, but I decided that old cities and towns almost anywhere in the world, which evolved organically following principles that Alexander tried to understand, are more beautiful and more functional than the ‘top-down’ approach of modern building design, even when this top-down approach is carried out in an inspired way, e.g. by Frank Lloyd Wright.
In future issues of This Week’s Finds I’ll interview Thomas Fischbacher about permaculture, and some of these ideas may come up.
It’s tricky to relate any of this stuff to mathematics — it’s possible that the mathematics governing the veins of a leaf is also lurking in the canals of Venice:
(Click to enlarge.)
• streamfortyseven says:
You might want to ask Dr Fischbacher if he’s ever heard of Dave Jacke, who wrote Edible Forest Gardens (see http://www.youtube.com/watch?v=ybhN0ep_0eE) which uses pattern language in permaculture design (see http://regenerativedesigns.files.wordpress.com/2010/04/fg_pl_sheets.pdf and http://appleseedpermaculture.com/forest-gardening-vision-pattern-language/)
btw, if the canals of Venice follow pre-existing watercourses, they most probably have fractal geometric character.
• John F says:
Among other references regarding fractal characteristics of rivers:
Nestler, J. M. and Sutton, V.K. (2000) “Describing Scales of Features in River Channels Using Fractal Geometry Concepts,” Regulated Rivers: Research and Management 16: 1-22
• Phil Henshaw says:
That’s interesting, as I recently found a link between Alexander’s language and mine. It’s that what he calls “harmony-seeking computation” is what I call “organizational learning process“. He pays more attention to spatial design features of the negotiation between the system and the environment and I key more on the eventful succession of change. Both involve a whole system forming as a successful way of facilitating innovative communication between the parts…
http://synapse9.com/blog/2011/02/27/the-fit-with-alexander-%e2%80%93-and-clearer-escape-from-our-traps/
3. Graham says:
It’s also possible that ‘green mathematics’ already exists in preliminary form, scattered throughout many different fields: mathematical biology, quantitative ecology, bioinformatics, artificial life studies, and so on. Maybe we just need more mathematicians to learn these fields and seek to synthesize them.
I think that is the case.
I mostly know about the ‘green’ maths that relates to phylogenetic analysis, and branching processes (trees not networks) though networks are increasingly being used too. Here’s a couple of articles whose titles at least seem likely to appeal to you.
• P D Jarvis, J D Bashford and J G Sumner, Path integral formulation and Feynman rules for phylogenetic branching models, http://iopscience.iop.org/0305-4470/38/44/002
• J. G. Sumner, B. H. Holland, P. D. Jarvis, The algebra of the general Markov model on phylogenetic trees and networks, http://arxiv.org/abs/1012.5165
• Frederik De Roo says:
The former article is also on the arxiv:
http://arxiv.org/abs/q-bio/0411047
• John Baez says:
Graham wrote:
I think that is the case.
Good. I hope so. It’s very hard, perhaps impossible, to invent a new idea from scratch. But mathematicians are good at taking vague glimmerings of new ideas and gradually making them more precise and more explicit.
Thanks for the references. It’s easy to drown under references in this game, so my thanks are not a covert way of asking other people to post more references—but thanks!
4. Ali Moharr says:
John,
Your article made me remember a beautiful ‘pattern’ I saw on Arthur Winfree’s interesting book, When Time Breaks Down. It describes the simulation of a 4D rotating scroll that is related to the electrochemical activity of heart muscles that drive the beating cycles of heart operation.
Speaking of analogies, and close similarities, I wonder if anyone can say something about Roger Penrose’s sketch of Robinson Congruence in his The Road to Reality, chapter 33, Fig 33.15.
Thanks,
Ali
5. John F says:
I don’t know anyone in bioinformatics using sbgn per se, even technophile early adopters. Just about everyone uses Python flavors of sbml,
http://sbml.org/SBML_Software_Guide/SBML_Software_Summary
or other Python network software
http://en.wikipedia.org/wiki/Genenetwork
6. Mark Meckes says:
Since you yearn for math about structures like this:
you might like to check out this paper:
• Qinglan Xia, The formation of a tree leaf.
(especially the figures on pages 13 and 14); here is a related talk:
• Qinglan Xia, The formation of a tree leaf, 9 November 2005.
The math used here is related to some of the work for which Cedric Villani recently won a Fields medal.
• Mark Meckes says:
Whoops, I tried to include a copy of the leaf image from your post, but it didn’t work, somewhat ruining my punchline.
• John Baez says:
I fixed that. Alas, WordPress has decided that I’m the only one able to post images on my blog. If anyone wants to post an image, just include the URL. If it’s pretty enough, and relevant enough, I can make it display on the blog.
Thanks for the leaf-related links. I am indeed curious about how plants create leaves (and many other things like this).
• streamfortyseven says:
Look at Barnsley’s Iterated Function Systems, too: http://www.superfractals.com/
• John Baez says:
Thanks, streamfortyseven. This brings back memories…
I spent some time thinking about iterated function systems back in early 1990s shortly after Barnsley first popularized them. In case anyone is wondering, iterated function systems are a simple and elegant way to generate fractals, some of which look suprisingly plant-like. The math is summarized nicely in Thomas Colthurst’s comment below. The example everyone loves is called Barnsley’s fern:
My friend Nate Osgood, then a computer science grad student at MIT, became curious when he heard that Barnsley had started a company, Iterated Systems, which claimed to do image compression using iterated function systems. In 1992 this company got a \$2.1 million government grant to work on this stuff, but the methods were secret—or at least we never heard anyone explain them. So, we tried to figure them out.
I wound up deciding that it wouldn’t really work very well. It’s impressive that the Barnsley fern can be encoded in just a few numbers using an iterated function system. But this is a very special image. If you take a typical image—like a human face, for example—I don’t think you can generate it efficiently using iterated function systems. And it’s even harder to imagine an automatic system where you hand it an image and it figures out how to compress it a lot using iterated function systems. I’d hoped that you could use some clever ideas involving harmonic analysis on the affine group, but after a while I decided not.
The subsequent history of attempts to do image compression using iterated function systems makes me think I was right. Read the Wikipedia article. You’ll see things like:
During the 1990s Iterated Systems Inc. and its partners expended considerable resources to bring fractal compression to video. While compression results were promising, computer hardware of that time lacked the processing power for fractal video compression to be practical beyond a few select usages. Up to 15 hours were required to compress a single minute of video.
So: that’s a summary of my old reasons for being less than thrilled with iterated function systems. My new reason is that I don’t believe Barnsley’s fern sheds a vast amount of light on how a fern actually grows.
It does shed a little bit of light. Briefly: if little bits of leaf keep growing littler bits following a simple pattern, you’ll get a fancy-looking leaf!
But then comes the question of how the fern actually does this. And here it’s worth noting that not all leaves are easily drawn using iterated function systems.
But now I’m curious about the state of the art, so now I want to read that paper by Qinglan Xia.
• Giampiero Campa says:
This is great stuff !
Perhaps it is obvious, but this also brings up the question about the right level of abstraction you want to model a given natural system.
Of course the “right” level of abstraction is the highest one that is able to answer your questions.
But i am afraid that for some natural systems you have to go way down the abstraction level to answer even basic questions. The weather and climate are an example.
I think that the leaf tree allows higher level description because it has evolved to do something specific in its environment, so to speak. So you might be able to model it without having to simulate the single atoms.
Sorry, i am basically just thinking out loud here, but perhaps it can spark an interesting discussion, or perhaps John will come up with another great link :)
• John Baez says:
Giampiero wrote:
I think that the leaf tree allows higher level description because it has evolved to do something specific in its environment, so to speak. So you might be able to model it without having to simulate the single atoms.
I don’t have anything exciting to say, but I think you’re right. It’s quite wonderful how evolution lets us apply the idea of final cause to biological systems. Aristotle had various concepts of cause, and one of them involved the notion of telos, or goal. For example, if you ask “why is a spoon concave?” it makes sense for me to say “in order for it to hold soup”, because it was designed to accomplish that goal. The marvelous thing is that even without ‘intelligent design’, evolution leads to the proliferation of entities whose structure can be understood (to some extent) by acting as if they were designed to achieve some goal!
So, for example: we can mathematically study a leaf, see what it would do to optimize something, and perhaps see that leaves in fact do this.
Physics on the other hand mainly deals with the concept of ‘efficient cause’, as in “the spoon is concave because you pushed down on it with a hard object”. Efficient causality is such a big deal in physics that this about the only kind of cause most physicists discuss when speaking of ‘causality’.
There is, however, one famous exception, namely that a particle tends to move along that path from start to finish that minimizes the action. This smells like a case of ‘final cause’, and it bugged physicists for a long time. Maupertuis, who came up with the principle of least action, wrote:
The laws of movement and of rest deduced from this principle being precisely the same as those observed in nature, we can admire the application of it to all phenomena. The movement of animals, the vegetative growth of plants … are only its consequences; and the spectacle of the universe becomes so much the grander, so much more beautiful, the worthier of its Author, when one knows that a small number of laws, most wisely established, suffice for all movements.
and there we see how the concept of ‘final cause’ tends to get linked to the concept of ‘intelligent design’.
But anyway, all this is just historical-philosophical chit-chat about junk I’ve known for ages. Qinglan Xia’s paper on the structure on the formation of a tree leaf is the sort of thing I’m actually interested in now.
• Florifulgurator says:
Thanks for the Xia link! Looks like the first convincing model of leaf growth. And it has ramifications to other (bio-)logistic systems. Thrilling. Green math indeed.
7. Peter Morgan says:
I very much look forward to what you have to say about categorification of Feynman diagrams, but I think of them as a shorthand that makes possible the extraction of global information from nonconvergent functional Taylor series, which places models into Universality classes. I wonder —although I admit this to be a prejudice— whether networks of all kinds are (only) essential waypoints towards understanding systems in qualitatively more global and continuous terms, which are, however, perhaps (only) a useful way to systematize discrete structures. There is always information that is not included in a finite network, though it is also doubtful to me —another prejudice— that any continuous structure would be more than a model.
I think you’ll recognize that what you are doing is “green mathematics” when it feels both holistic and mathematics. [I'll be back to earth later today. You and Tim van Beek together sent me into orbit.]
• Peter Morgan says:
I know the above is off the wall, but this afternoon I came across a comment on the website of the Courant Research Center in Göttingen that says something like it: Today we live in a period during which very different areas of mathematics come closer together and exchange techniques and ideas. In this process, new problems occur, e.g. if the “flexible” world of topology and geometry is used in the “rigid” world of number theory. For the time being, we are still lacking a good understanding of the overarching structure which makes this efficiently possible.
• John Baez says:
Peter wrote:
I very much look forward to what you have to say about categorification of Feynman diagrams.
You can read about that here:
• John Baez and James Dolan, From finite sets to Feynman diagrams, in Mathematics Unlimited – 2001 and Beyond, vol. 1, edited by Björn Engquist and Wilfried Schmid, Springer, Berlin, 2001, pp. 29-50.
and in a much more gentle expository way in these course notes:
•Quantization and categorification seminar: Fall 2003, Winter 2004 and Spring 2004.
and then more formally, with new ideas added, in this paper by a student who took that course:
• Jeffrey Morton, Categorified algebra and quantum mechanics, Theory and Applications of Categories 16 (2006), 785-854.
This is the some of the ‘old stuff I used to love’. But now Jacob and I are adapting it for applications to chemistry and population biology.
I’ll probably focus on the new aspects, and try to write about them in a way that doesn’t assume folks know the old stuff.
I think you’ll recognize that what you are doing is “green mathematics” when it feels both holistic and mathematics.
I think you’re right. I’m not sure “holistic” is quite the word for it, but there’s a kind of feeling, almost like a taste in my mouth, that I get sometimes when things seem to be going in the direction of “green mathematics”. It’s sort of weird. But a lot of science starts from intuitions like this, just as much as on logic or experiment. Only time will tell if it’s a good intuition or a delusion.
[I'll be back to earth later today. You and Tim van Beek together sent me into orbit.]
Heh. I’m glad you find it exciting!
8. Thomas Colthurst says:
After that beautiful picture of a leaf, I was really expecting something about the mathematics of fractals. (http://en.wikipedia.org/wiki/File:Fractal_fern_explained.png, for example).
But in the spirit of John’s post let me tell you how to get a graph from an iterated function system.
An iterated function system or IFS is just a compact set of contractive mappings on a complete metric space. Every IFS has an unique attractor $A$ satisfying
$A = \bigcup_{f \in IFS} f(A).$
Define the graph of an IFS by having a node for each mapping and an edge between $f_1$ and $f_2$ iff $f_1(A)$ and $f_2(A)$ intersect.
The fun thing is that the graph of an IFS tells you a lot about the attractor. For example, $A$ is connected iff the graph is. If the graph has no edges, then $A$ is Cantor dust (except in the one special case when all the mappings contract to a single point).
• John Baez says:
Thanks for explaining the idea! Despite my skepticism about applications of iterated function systems to biology and image compression, the math is definitely neat.
9. Krzysztof says:
chemistry, biology, ecology, statistics, and so on. There should be a mathematical theory underlying the use of networks in all these disciplines.
Algebraic geometry? “Graphical models” (bayesian networks, hidden Markov models) are algebraic varieties!
Bernd Sturmfels studies combinatorial independence (previously he was into matroids) and seems to lead a programme of applying algebraic geometry of toric varieties to various problems in statistics, biostatistics (phylogenetics included) and machine learning.
• John Baez says:
Krysztof wrote:
Algebraic geometry?
Algebraic geometry by itself isn’t particularly ‘green’. But it’s a tool that you gotta know if you want to do mathematics without a limp. So yes: we’ll probably be seeing it, along with many other bits of math we know and love, in ‘green mathematics’.
“Graphical models” (bayesian networks, hidden Markov models) are algebraic varieties!
How does a Bayesian network give an algebraic variety?
Bernd Sturmfels studies combinatorial independence (previously he was into matroids) and seems to lead a programme of applying algebraic geometry of toric varieties to various problems in statistics, biostatistics (phylogenetics included) and machine learning.
Hmm! Thanks for telling me!
It’s sort of a side-issue in my current plan, but sometimes I’ve imagined talking about the relation between Petri nets and toric varieties. If I don’t ever get around to it, interested people will just have to read these:
• Craciun, Dickenstein, Shiu and Sturmfels, Toric dynamical systems.
* Leonard Adleman, Manoj Gopalkrishnan, Ming-Deh Huang, Pablo Moisset and Dustin Reishus, On the mathematics of the law of mass action.
I knew about these papers, but I didn’t know Sturmfels has a whole “programme” for applying toric varieties to different problems.
• Krzysztof says:
Sorry for I have no time for a proper response.
How does a Bayesian network give an algebraic variety?
Like this
• Luis David Garcia, Michael Stillman, Bernd Sturmfels, Algebraic Geometry of Bayesian Networks (+ accompanying thesis).
Well, I lied slightly about what, exactly, gives what ;)
There are books; electronic:
• Mathias Drton, Bernd Sturmfels and Seth Sullivant, Lectures on Algebraic Statistics
And dead tree, quite green nevertheless:
• Lior Pachter and Bernd Sturmfels, Algebraic Statistics for Computational Biology.
It is also helpful (almost like in “assumed”-helpful (that is, besides the whole intimidating lot of assumed AG)) to know about uses of graphical models in statistics.
Presumably, there are authors better suited to our “networks” topic but I also wanted to point at the (unrelated) concept of “variational inference” that Michael I. Jordan pushes:
• Michael I. Jordan, Graphical Models (overview).
• Martin J. Wainwright and Michael I. Jordan, Graphical Models, Exponential Families, and Variational Inference (book, 300 pp.)
• John Baez says:
Thanks for all these references, Krzysztof — they look quite relevant! Someday I hope to explain some of my ideas about those “graphical models” you mention above. I hadn’t seen any connection between them and, say, toric varieties (which I see coming up in Petri net theory). I still don’t see the connection but it sounds like you’re hinting at one. I’ll have to read some of this stuff.
10. Krzysztof says:
Also, information geometry too admits a natural description within (real) algebraic geometry. There’s a (involved) book “Algebraic Geometry and Statistical Learning Theory” by Sumio Watanabe which beyond above also develops (not terribly practical at the moment) methods for graphical models from that viewpoint.
11. John F says:
I have a friend who is doing his thesis on a tree. Specifically it’s a green Mechanical Engineering Ph.D. about transport processes for a single tree in situ, with hundreds of sensors I helped install. Some of these are soil microtensiometers. Google tensiometry and transpirational pull http://en.wikipedia.org/wiki/Transpirational_pull
if you’re looking for some weekend escapism.
• John Baez says:
John F wrote:
I have a friend who is doing his thesis on a tree.
For a second I thought he was trying to outdo Julia Butterfly Hill!
Google tensiometry and transpirational pull http://en.wikipedia.org/wiki/Transpirational_pull
if you’re looking for some weekend escapism.
Sounds interesting—thanks!
12. Steven Colyer says:
…mathematicians are especially good at soaring above the particulars and seeing general patterns.
Right on! That is to say: abstracting. Topology is beautiful, and once again, we see the name Euler pop up in its beginnings, first in his lovely V-F+E = 2 + 2r to describe polyhedra (actually minus the 2r if a convex polyhedra) Polyhedron formula, then again in attacking the Bridges of Konigsberg (now Kalingrad) problem, which I do believe was the start of Network theory, yes?
In any event, just last week I learned that Network theory (as boyhood love of mine) had been renamed Graph Theory, which means I have to go back and study Fotini’s stuff!
The Rosetta Stone paper you and Mike Stay have written looks sweet as well, and I look forward and to reading it and calling attention to it soon, so thanks to you and Mike.
Yes, we live in interdisciplinary times. In 1950 Math and Physics were never father apart, and logic-based Computer Science was just getting started. 61 years later we see how interconnected they are, and props to you and all others making that point. good luck, sirs. And ladies.
13. Phil Henshaw says:
The origin of my strange approach to “green mathematics” was being unable to conceive of any kind of equation that would be able to be responsive to its interactions with its environment, the way physical systems do.
So that’s why I invented my “other mathematics” which is more of a carefully constructed “pattern language”, to serve as an environment that equations could potentially relate to. As an “artificial environment” it describes boundary conditions within which various kinds of developmental progressions might take place.
So far there are only very limited features of that relationship that I can see how to automate, except for the approach of natural boundaries posing questions about how to change the equations. In that way it seems to naturally become an aid for people using a combination of equations and observations, as a way to learn about how defined and undefined systems are interacting.
Has anyone else found any other way to emulate the back and forth conversation between systems and natural environments, that I might find useful?
• Uwe Stroinski says:
You could consider non-axiomatic foundations of mathematics. In such frameworks axioms (as a priori true statements) are replaced by relations. The truth of assertions can only be described in relation to the truth of other assertions. One can then reason about feedback loops and other circular systems in a systematic way. The price to pay is a higher vagueness of the results due to the multi-valuedness of the logic. As far as I know, there is no active research in this area. Maybe for a reason, since there are no relevant applications to traditional mathematical topics like number theory and one has to give up a lot (like the notion of proof becoming different).
While writing this I realize, that this is a contender for John’s ‘green mathematics’ with a ‘radical shift of viewpoint’.
• Phil Henshaw says:
Uwe, Yes, I inject fuzzy relationships at strategic points in the relation between models and their environments. That comes out of looking at how far from normal a system might be able to diverge. That serves to hypothesize a zone within which a model is self-contained, and beyond which interactions with other things not in the model arise at some point. It’s how I get questions about the systems’s environment into the model, by anticipating the onset of abnormal behavior rather than seeing uncertainty as only the limits of normal behavior.
It’s true that lots of people and a lot of talent went into exploring fuzzy logic and vague relationships for building logical networks. I don’t know where that ended up really, but it doesn’t seem to have fulfilled it’s once great promise. Failed experiments often do produce at least a few bits of learning that one can use with high confidence, though.
You might have a short list of those, but one seems to be that mathematics doesn’t seem to work well with undefined variables, or ones that essentially say “something will upset the model”, “go look at your environment”, but when studying open systems that’s exactly what we’re trying to model.
When you look at it, how physical systems interact with their environments is not definable for a host of interesting reasons. One is that environments can support many different kinds of systems all of which located by their internal consistency and mutual independence. So how they would interact would not be definable withing the logic of either. To me that was a thrilling discovery, a new way to look at the problem.
14. Ali Moharrer says:
Kent Palmer is a sociologist-philosopher who has written extensively on the concepts of ‘emergence’ and complex systems, mainly based on general systems science and its extension to meta-systems. I have been studying his 2nd doctoral thesis that talks about his investigations into the roots of Emergent Design. You might want to look him up. Check this link:
http://works.bepress.com/kent_palmer/
15. [...] So you cheered on Math4love for the Monthly Math Hours, followed the intriguing discussion of Network Theory and Green Mathematics at John Baez’s Azimuth, were amazed at the Math History Tour of Nottingham with your friendly [...]
16. John F says:
Not necessarily very green, but the two big commercial operations research and network analysis software systems are the unimaginately named Netica
http://www.norsys.com/
and Analytica
http://www.lumina.com/
17. Marc Manley says:
I wish there were a branch of mathematics—in my dreams I call it green mathematics—that would interact with biology and ecology just as fruitfully as traditional mathematics interacts with physics.
I had a thought here (and I know I’m swimming in deep waters), but perhaps your wish might be fulfilled if you thought about “greening” from the “O” part instead of the “I” part of I/O. In a sense, so long as your mathematical output results in something green, this would relieve the necessity of creating some sort of “green math”. Math is math. Let the results of your math be green and there you have it. Just a thought. Good article.
• Phil Henshaw says:
One of the catches for either an input or output side analysis is that all equations operate in an open environment. Other business models, also constructed around their own I/O equations, will be “out there” to interact with. Neither’s equations will contain information about their future interactions with others. That’s the problem of open environments, and a daunting challenge for using “stand-alone math” for modeling their parts independent of the whole.
The available solution isn’t perfect, of course, but a great many of the important interactions that develop over time can be identified from closely observing how the larger system behaves as a whole. That way mathematical models, unable to define their own environment, can still interact with their environment through the critical observations of the model maker.
For example, one might be concerned with the major new wealth transfer presently occurring from the physical resource using side of business to the largely non-resource using side largely using financial information as a business. Think of the enormous resource price escalation that began around 2003 as a wholesale transfer of working capital from one side of the economy to the other.
It seems to be for natural causes, that global demand is exceeding global supply for the whole network of formerly interchangeable food and fuel resources. The price jump appears to be caused by finance responding by setting prices according to scarcity rather than material cost. It also seems likely to be permanent, till something else reverses the trend of increasing resource demand.
The point is that this kind of change of environment, for “green” products, is bad for business. It would cause struggling business plans to fail first, perhaps costing the economy much of the present wave of creative start-ups that are trying to invent our way out of the natural limits crisis we’re confronting.
See how that works? It’s important to inventors to know how their environment is changing. Stand-alone formulas can’t exist for how a formula’s assumptions need to change.
Here is a page of graphs from 1950 to the present showing what seem like the key resource market factors in our present crunch. I assembled them from WorldWatch Vital Signs 2011.
http://www.synapse9.com/issues/PlanteChange11_03.pdf
18. streamfortyseven says:
For large networks:
The map equation
M. Rosvall, D. Axelsson, and C. T. Bergstrom
European Journal of Physics 178:13-23. Also arXiv:0906.1405v1 [physics.soc-ph]
(
“Networks are useful constructs to schematize the organization of interactions in social and biological systems. Networks are particularly valuable for characterizing interdependent interactions, where the interaction between components A and B influences the interaction between components B and C, and so on. For most such integrated systems, it is a flow of some entity – passengers traveling among airports, money transferred among banks, gossip exchanged among friends, signals transmitted in the brain – that connects a system’s components and generates their interdependence. Network structures constrain these flows. Therefore, understanding the behavior of integrated systems at the macro-level is not possible without comprehending the network structure with respect to the flow, the dynamics on the network.”
(http://octavia.zoology.washington.edu/publications/RosvallEtAl10.pdf)
Multilevel compression of random walks on networks reveals hierarchical organization in large integrated systems
M. Rosvall and C. T. Bergstrom
arXiv:1010.0431
“To comprehend the hierarchical organization of large integrated systems, we introduce the hierarchical map equation that reveals multilevel structures in networks. In this information-theoretic approach, we exploit the duality between compression and pattern detection; by compressing a description of a random walker as a proxy for real flow on a network, we find regularities in the network that induce this system-wide flow. Finding the shortest multilevel description of the random walker therefore gives us the best hierarchical clustering of the network — the optimal number of levels and modular partition at each level — with respect to the dynamics on the network”
(http://octavia.zoology.washington.edu/publications/working/RosvallAndBergstrom10.pdf)
More available in PDF at:
http://octavia.zoology.washington.edu/publications/publications.html
19. [...] Network Theory (Part 1) [...]
20. [...] c’est ici [...]
21. Andrew Palfreyman says:
And then there’s fuzzy logic! And other assorted paradigms from AI (agents, expert systems, cellular automata and so on).
Reverse engineering real cellular networks is a worthwhile pursuit also. Knowing the circuit diagram of a cell allows deep, “bottom up” comprehension / simulation of “everything in biology” :). Sound good?
So – if I present you with a time series of matrices of interactions between N components, can you evince the “circuit diagram”? Even if it contains positive and negative feedback loops? Even if they are nested to arbitrary depth? (this sounds like the kind of puzzle that “Dr. Ecco” (Dennis Shasha) would set.
22. Ben says:
Hi Again,
If I am not mistaken, by a green mathematics, you mean a foundation for mathematics which can be seen as a “way of reasoning” that is also “green”. Category theory as a foundation is a subject in the philosophy of mathematics. The problem is, we are always presented the theory of categories in SET. Thus, where is the foundation?
If you look at the diagrammatic calculus of Coecke and Penrose, and especially the paper by Joyal, you see that the reasoning is primarily the reasoning about directed graphs. But every category is a directed graph with algebraic data over the edges. There is a relationship (which I am not entirely clear on) between the axioms of these diagrams and the axioms of what is called “linear logic”. Thus, given your interest in diagrammatic reasoning, perhaps the “Green Mathematics” is simply the presentation of the theory of categories in a linear logic with all axioms supported in the rewrite rules of a diagrammatic calculus. As for why this might be green, I have a blog post about how autocatalytic reactions (prominent in the molecular theories of life) are traces in a symmetric monoidal category:
http://whyilovephysics.blogspot.com/
Thus, we probably want our “Green” math to be a foundation with traces. Like I’ve said, this is a presentation of the theory of categories in a linear logic.
I’m a big fan of John’s work and this blog is especially interesting.
Best.
• John Baez says:
I’m glad you like my stuff, including this blog.
This blog post is the first of a series, on network theory. You can see the whole series—or at least the part I’ve written so far—here. There’s a lot more yet to come. In the long run it will connect with the diagrammatic calculi used by Penrose, Abramsky–Coecke, Joyal–Street and others.
There is a relationship (which I am not entirely clear on) between the axioms of these diagrams and the axioms of what is called “linear logic”.
I was confused by this relationship for a long time. Right now my best understanding can be found in a paper with Mike Stay, in the second to last section. We explain the relation between string diagrams and multiplicative intuitionistic linear logic, or MILL. This is a simple, easily comprehensible fragment of linear logic.
I have a blog post about how autocatalytic reactions (prominent in the molecular theories of life) are traces in a symmetric monoidal category…
Thanks, I’ll read that soon!
Computer scientists have long been interested in the relation between traces and feedback, and there’s a nice account of these ideas in Joyal and Street’s paper on traced monoidal categories. Since autocatalysis is a form of feedback, I guess you’re talking about something similar.
If I am not mistaken, by a green mathematics, you mean a foundation for mathematics which can be seen as a “way of reasoning” that is also “green”.
That’s a very ambitious goal. Usually the so-called ‘foundations’ of a subject are only established fairly late, when people have come to settled opinions about what they’re doing. So, instead of trying to figure out the foundations, I’m starting by working out the relationship between a bunch of existing approaches to complex systems made of interacting parts.
• Ben says:
I am not sure if I am interested in a foundation or not. As a scientist, I want to have direct, intuitive access to the underlying reasoning of the mathematical structure which I will then use to describe my contact with nature. Furthermore, I want to see the most basic aspects of that structure well reflected in the visceral experience of observation. To have this, I’ve found I’ve had to touch on foundations simply because a theory, like the theory of categories, is always presented in some ambient reasoning structure. I want direct access, I suppose.
I think that categories, are a great start to a green mathematics. For instance, I have a dream of modelling economic growth with technological growth built right in at the bottom. This kind of growth is “network diagram growth”. What I mean is the following. When we draw a diagram, we start with a single line. Then we draw a dot. Then we draw another line. If we slow this process down and do it in stages, we have three diagrams. A dot, a line with a dot, and a line a dot and a line. Thus, drawing a diagram is the process of diagram growth.
This growth is highly structured. It is similar to Sorkin’s Causal growth dynamics and it is similar to Panangaden and Martin’s Domains as spacetimes (since the domain map is the evolving causal structure). It is different, in that the basic thing that is growing is a graph, not a partial order.
I have attempted to give precise rules for this growth. This has lead me to think about continuous functors, since the diagrams encode the axioms of particular categories.
In any case, “network diagram growth” is a technological growth,in that, a small diagram can be seen as a small category: one with very little going on. It’s like a factory that does one thing like grinding wheat into flour. Technological growth, happens when the factory owner realizes he can mix flour and water and then bake bread. That produces a new diagram and the old one maps into it in a structure preserving way. In this light, technology is understood as “learning about your past”. The larger diagram can be seen as a context in which to interpret one’s past.
This is “green”, in that, it is similar to how an organism might grow and change as when a seed grows into a tree. It is also “green”, in that, we now have a basic way to understand the value of economic growth. Namely, it allows us to understand our past. In that way, it is truly remarkable. Also, if we have good mathematical models of growth, perhaps we can have growth that is healthier for the planet. We can direct it at evolving the species, rather than just increasing the gross amount of resource we are consuming.
23. Ben says:
I should clarify what I mean by a healthier growth.
A factory has some material inputs and processes them, using process A, into some material outputs. The owner skims a profit off the top. If he increases the amount of throughput, ie the number of times that process A is enacted per day, then he grows his profit. This is one of the dimensions of growth.
The other dimension of growth is where the owner of the plant invents a new process, B, which takes some of the original inputs and produces new outputs. This new process has higher value than the last one (like they way new technologies fetch a higher price). This is growing value. He can take some of the material that was originally allotted to process A and use it to start enacting process B. Thus, the total amount of resource is the same, but because B fetches a higher price, his bottom line grows.
This is how we can have growth, without increases in gross consumption.
24. Jeff Broadbent says:
Great stuff but you have skipped over social networks. We are using these to collect data on differing societal responses to climate change. The project is described at our website http://www.compon.org. It would be great to have some nat sci math networks input
25. [...] interesting programming puzzle to chew on: look over John Baez’s posts on Network Theory. Learn some stuff on Petri nets (see also Azimuth project wiki on Petri [...]
26. Arjun Jain says:
Inspired by your blog, I started one a few days ago.
I don’t know if you knew of this before, but you might like this:
http://arjunjainblog.wordpress.com/2013/04/14/phyllotaxy-a-serendipitous-surprise/
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## Is R^3 the square of some topological space?
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The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected.
A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this.
So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$?
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I'm wondering to what extent there is unique factorization of topological spaces relative to $\times$. $\mathbb{Q}$ is an idempotent (as is its complement in $\mathbb{R}$), but are there more interesting failures of UF involving connected spaces? Or results establishing UF for "nice" families of spaces? Should these be posted as a new question? – Yaakov Baruch Apr 3 2011 at 1:41
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Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? – Yaakov Baruch Apr 4 2011 at 16:38
Without knowing any algebraic topology, it's possible to conclude at least something about X. If X is metric, compact, or locally compact and paracompact, then $\dim(X\times X)\le 2\dim X$, which means X has to have Lebesgue covering dimension at least 2. Wage, Proc. Natl. Acad. Sci. USA 75 (1978) 4671 , www.pnas.org/content/75/10/4671.full.pdf . What is the weakest condition that guarantees $\dim(X\times Y)= \dim X+\dim Y$? Given Yaakov Baruch's comment about the "dogbone space," it's not obvious that X is at all well behaved simply from the requirement that its square is $\mathbb{R}^3$. – Ben Crowell Jan 19 at 15:55
## 3 Answers
No such space exists. Even better, let's generalize your proof by converting information about path components into homology groups.
For an open inclusion of spaces `$X \setminus \{p\} \subset X$` and a field $k$, we have isomorphisms (the relative Kunneth formula) ```$$
H_n(X \times X, X \times X \setminus \{(p,p)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{p\};k) \otimes_k H_q(X, X \setminus \{p\};k).
$$``` If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial. However, if `$H_p(X, X \setminus \{p\};k)$` were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.
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I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. – Georges Elencwajg Apr 2 2011 at 19:40
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this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.
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I hope no one misses this nice alternative proof because it's behind a link. – Richard Dore Apr 4 2011 at 2:24
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Quoting from the link: "The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E4", R.H. Bing, Annals of Mathematics series 2 vol 70 1959 pp. 399–412) which constructs an extremely pathological space B, called the "dogbone space", not even a manifold, which nevertheless has B × R^3 = R4." This is relevant to my comment to the OP. – Yaakov Baruch Apr 4 2011 at 5:16
I don't understand this step in the proof: Why does the map $X^4 \to X^4, (a,b,c,d) \mapsto (c,d,a,b)$ correspond to the map $R^6 \to R^6, (p,q,r,s,t,u) \mapsto (s,t,u,p,q,r)$? I mean, the homeomorphism is not supposed to commute with projections ... – Martin Brandenburg Apr 4 2011 at 15:05
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@Martin: The homeomorphism $(X\times X)\times (X\times X)\cong \mathbb R^3 \times \mathbb R^3$ respects projections by construction, so swapping the "two factors" (which I've emphasized with parentheses) on the left hand side corresponds to swapping the two factors on the right hand side. – Anton Geraschenko♦ Apr 5 2011 at 5:42
I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.
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Would you care to elaborate? – Agol Jan 19 at 5:09
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All things considered, perhaps "S" is not the best name for the topological space for this assertion. – Terry Tao Jan 19 at 5:52
@Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. – Adam Epstein Jan 19 at 10:28
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@Agol Fix $s\in S$. On the one hand, $\pi_2(S\times S,(s,s))\cong \pi_2(S,s)\times\pi_2(S,s)$. On the other hand, $\pi_2({\bf S},{\bf s})\cong{\mathbb Z}$ for any 2-sphere $\bf S$ and any ${\bf s}\in{\bf S}$. Now it suffices to observe that ${\mathbb Z}\not\cong G\times G$ for any group $G$: indeed, such a group must be an infinite quotient of $\mathbb Z$, whence $G\cong{\mathbb Z}$, but ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ – Adam Epstein Jan 19 at 11:27
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As it happens, this started out as a wry comment about a different post, namely mathoverflow.net/questions/115799…. Then I noticed this question and accidentally posted as an answer what was intended as a mere comment. – Adam Epstein Jan 19 at 16:59
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http://math.stackexchange.com/questions/258332/prove-that-if-ax-by-abxy-then-xy-1?answertab=votes
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# Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.
Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.
How do I use logarithms to approach this problem? Any help would be appreciated, thanks.
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## 4 Answers
Using logarithms:
Since $a^x = b^y$, $$\log a^x = \log b^y \quad \Rightarrow \quad x \log a = y \log b \quad \Rightarrow \quad \log a = \frac{y}{x} \log b$$ Then, since $b^y = (ab)^{xy}$, $$\log b^y = \log (ab)^{xy} \quad \Rightarrow \quad y \log b = xy \log (ab) = xy \left( \log a + \log b\right)$$ Let's assume $y \neq 0$ (since if $y=0$, then we must also have $x=0$ and get the required conditions without having $x+y=1$ -- meaning the original question must have had some restriction such as $x, y \neq 0$). Cancel $y$ on both sides of the last equation: $$\log b = x\left(\log a + \log b\right) = x\left( \frac{y}{x} \log b + \log b \right) = y \log b + x \log b = (y + x)\log b$$ Then as long as $b \neq 1$ (which is guaranteed if $x \neq 0$), we know that $\log b \neq 0$, hence we can cancel in the equation: $$\log b = (x+y)\log b \quad \Rightarrow \quad 1 = x + y.$$
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No logarithms are needed:
$$a^x=(ab)^{xy}=a^{xy}b^{xy}=\left(a^x\right)^y\left(b^y\right)^x=\left(a^x\right)^y\left(a^x\right)^x=\left(a^x\right)^{x+y}$$
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Hm, I wouldn't be surprised if I'm missing something obvious here, but how do you get $(a^x)^1 = (a^x)^{x+y}$ implies $1 = x+y$ without logarithms? In other words, how can we get this result without implicitly taking a logarithm while stating that $x \to a^x$ is injective? Also, I think the case $a = b = 1$ has to be excluded. – brom Dec 14 '12 at 1:14
@brom: Depends on the order in which one defines and proves things about exponentials and logs. It’s quite possible to define the exponential functions and prove that they’re injective (for base not equal $1$, of course) before dealing with logs at all. – Brian M. Scott Dec 14 '12 at 1:19
Indeed $a=b=1$ has to be excluded. $f(x)=a^x$ is injective provided $a > 0$ and $a \neq 1$ because this function is monotone increasing for $a > 1$ and monotone decreasing for $a < 1$. – proximal Dec 14 '12 at 1:20
@BrianM.Scott What about Hugo's response? – Alan Dec 14 '12 at 4:19
@Alan: What about it? It’s perfectly correct. Mine was never intended to be a complete answer: I was addressing the main case, presumably the one of greatest interest to the OP. – Brian M. Scott Dec 14 '12 at 4:46
show 9 more comments
How about $x=y=0$ ? Am I missing something?
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You missed $a=b=1$, or $a=b=0$, or $a=b=-1$ with $x$ and $y$ both even integers – Henry Dec 14 '12 at 7:21
How about this?$$\begin{align} &(ab)^{xy} \\ =& a^{xy}\cdot b^{xy} \\ = & (a^x)^y \cdot (b^y)^x \\ = & (a^x)^y \cdot (a^x)^x \\ = & (a^x)^{x + y} \end{align}$$It suffices to say that $x + y = 1.$
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http://mathhelpforum.com/calculus/22650-integral-problem.html
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# Thread:
1. ## Integral problem
Find f(x) if f(1) = -1 and the tangent line at (x, f(x)) has slope 2e^x + 1
This is a problem that was thrown into my indefinite integral homework, and I'm kinda stuck.
any help appreciated.
2. Originally Posted by leviathanwave
Find f(x) if f(1) = -1 and the tangent line at (x, f(x)) has slope 2e^x + 1
This is a problem that was thrown into my indefinite integral homework, and I'm kinda stuck.
any help appreciated.
the derivative gives the slope. thus if the slope at any $x$ is $2e^x + 1$ it means that:
$f'(x) = 2e^x + 1$
$\Rightarrow f(x) = \int f'(x)~dx$
and use the fact that $f(1) = -1$ to solve for the arbitrary constant of integration
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http://math.stackexchange.com/questions/188001/membership-based-on-maximum-of-a-function-over-the-set?answertab=oldest
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# Membership based on maximum of a function over the set.
Let $S\subset \mathbb{R}^n$ and let $f(x)$ be a continuous function over $\mathbb{R}^n$. Furthermore, define $s_{\text{max}}:= \sup_{x\in S} \{f(x)\}$ and let $f(x)$ attain its minimum for at least one element in $S$.
Under what conditions on $f(x)$ and $S$ does it hold that $f(x_1)\leq s_{\text{max}}$ $\Rightarrow$ $x_1\in S$?
In particular, does this hold if $f(x)$ is a convex function and $S$ is a convex set? If $f(x)$ is convex and $S$ is an arbitrary set?
Unfortunately I am far from my comfort zone here, so any help is much appreciated!
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Do you mean $f(x) < s_{max} \implies x \in S$ or $f(x) < s_{max} \Leftrightarrow x \in S$? In the latter case, no such function exists as then $x_{max} \notin S$ and in the former, for any set any constant function works. You might have wanted $\leq$ instead. Also, some sets don't have a $\max$, you might want to use $\sup$ instead. – Karolis Juodelė Aug 28 '12 at 17:27
I am only interested in $f(x)\leq s_{max} \Rightarrow x\in S$. Thanks for your comments, I have amended the question based on them. Yes, a constant function would work but I would like to know what other possibilities are allowed. – Juan Miguel Arrazola Aug 28 '12 at 18:43
## 3 Answers
Let $\Sigma_S = \{x | f(x) \leq \sup_{t \in S} f(t) \}$. If $x \in S$, then we have $f(x) \leq \sup_{t \in S} f(t)$, hence $x \in \Sigma_S$, i.e, $S \subset \Sigma_S$. This is true for all $S,f$.
The desired property above is equivalent to $\Sigma_S \subset S$, hence this is true iff $S = \Sigma_S$.
That is, the property you desire is true iff $S$ has the form $\{x | f(x) \leq L \}$ for some $L$ (which may be $\infty$).
For a counterexample, choose the convex function $f(x) = x^2$ on the convex set $[0,1]$. It is easy to see that $s_\max =1$, but $f(-1) \leq 1$ also.
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This is precisely what I was looking for. Many thanks! – Juan Miguel Arrazola Aug 28 '12 at 22:32
Let's assume that $S$ is a subset of the domain of $f$ which -- for the case of maximisation -- we define as $\operatorname{dom}f=\{x|f(x)>-\infty\}$. Then, the fact that $f(x)\leq s_{max}$ does not imply that $x\in S$ nor that $x\notin S$. You can see this through a simple example. Consider the concave function $f(x)=-x^2$ and the convex set $S=[-1,1]$. Then $s_{max}=\sup_{-1\leq x \leq 1}-x^2=0$ and take $x_0=2\notin S$. Notice that $f(x_0)\leq s_{max}$. You can easily find counter-examples for your statement using convex functions. Take for instance the convex function $f(x)=x^2$ and $S=[1,2]$ and $x_0=0$.
One useful things about convex minimisation over convex sets, is that any local minimum you find is also global. The same holds true for maximisation problems when the function is concave and the set convex.
I can only think of an extremely special case where your statement holds true. $f:\mathbb{R}^n\to\mathbb{R}$ is quadratic, i.e. $f(x)=x'Qx$ for some positive definite matrix $Q$, and $S$ is of the form $S=\{x\in\mathbb{R}^n|x'Px\leq \gamma\}$ for some $\gamma>0$ and $P=P'$ positive definite.
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I'll try answering the following question. What kind of set $S \subset \mathbb{R}^n$ can be defined as $\{x|f(x) \leq c\}$ for a continuous $f : \mathbb{R}^n \rightarrow \mathbb{R}$ and constant $c$.
Trivially, the function $f(x) = \inf_{y \in S} d(x, y)$ - distance from $S$ to $x$ defines any closed set $S$. In general the shape of $f$ can vary a lot as long as $f(\partial S) = c$ and $f$ has the right relation to $c$ in the interior and exterior of $S$. Open sets would work if the condition was $f(x) < c$. Now $f$ would measure the (negative of) distance from $x$ to the complement of $S$ instead.
This does not cover all sets though. $\mathbb{Q} \subset \mathbb{R}$, for example is neither closed nor open. It is clear that unless $f(\mathbb{Q}) = c$ the function will not be continuous. Now take any Cauchy sequence $(x_n)$ that does not converge in $\mathbb{Q}$. Due to continuity of $f$ we have that $f(\lim x_n) = \lim f(x_n) = \lim c = c$ - a contradiction.
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http://mathhelpforum.com/advanced-applied-math/180135-ivp-using-leibniz-formula.html
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# Thread:
1. ## IVP using Leibniz formula
How does one formulate this integral equation as an initial value problem?
u(t)=1 + integral from 0 to t (s ln (s/t) u(s) ds)
we're supposed to use Leibniz formula but I'm not sure how to apply it. Thanks!
2. So your integral equation is
$u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$
and you need to convert to an IVP using the Leibniz formula, which is, in this case,
$\frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$
So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?
3. Originally Posted by Ackbeet
$u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$
and you need to convert to an IVP using the Leibniz formula, which is, in this case,
$\frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$
So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?
Why not just note that $\displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?
4. Originally Posted by Drexel28
Why not just note that $\displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?
True enough, but if the OP'er must use the Leibniz formula because of class restraints, then we're back to square one.
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http://unapologetic.wordpress.com/2007/08/15/an-example-of-an-enriched-category/?like=1&_wpnonce=274c59a080
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# The Unapologetic Mathematician
## An example of an enriched category
Sorry for the delay, but the cable setup took more than I’d expected (there will me more on this over at Yankee Freak-Out). Today, I’d like to run through an example of a monoidal category, and what sort of enriched categories it gives rise to.
The category I’m interested in is the ordinal $\mathbf{2}$. Remember that this consists of the objects ${0}$ and $1$, with one non-identity arrow $0\rightarrow1$. We can make this into a monoidal category by saying $a\otimes b=ab$. Then $1$ is the monoidal identity object.
So what is a category enriched over $\mathbf{2}$? Well, first it has a collection of objects. For each pair $(A,B)$ of objects we either have the hom-object $\hom_\mathcal{C}(A,B)=0$ or $\hom_\mathcal{C}(A,B)=1$.
To have “identity morphisms” means we need an arrow $1\rightarrow\hom_\mathcal{C}(C,C)$ for each object $C$. But the only such arrow in $\mathbf{2}$ is $1\rightarrow1$, so $\hom_\mathcal{C}(C,C)=1$. For composition, we need arrows $\hom_\mathcal{C}(B,C)\otimes\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(A,C)$. Thus if $\hom_\mathcal{C}(A,B)$ and $\hom_\mathcal{C}(B,C)$ are both $1$, then so must be $\hom_\mathcal{C}(A,C)$.
Now we can see that this is just a different way of talking about a preorder. The identity morphism corresponds to the reflexive axiom, and the composition morphism corresponds to the transitive action. In short: $A\preceq B$ if and only if $\hom_\mathcal{C}(A,B)=1$.
Another example I’ve seen bandied about uses the category $\mathbf{pSet}$ of “pointed sets”. This is just a set with an identified “point”. For example, $(\{1,2,3\},1)$ is a pointed set, and $(\{1,2,3\},2)$ is a different pointed set. The morphisms are “pointed functions”, which have to preserve the point — a morphism $f:(X,x_0)\rightarrow(Y,y_0)$ consists of a function $f:X\rightarrow Y$ with $f(x_0)=y_0$. This category has finite products, so it’s monoidal.
The usual statement is that categories enriched over $\mathbf{pSet}$ are the same as categories with “zero morphisms”. These are like categories with zero objects, but without needing an object to factor things through. Every hom-set has a special “zero” morphism, and the composition of a zero morphism with any other morphism is another zero morphism. The problem is, whenever I try to show this it doesn’t seem to work out. I think that there’s something askew or oversimplified with the statement somewhere.
Your mission, should you choose to accept it, is to figure out what the right statement is, and to prove it. Let me know by email (if you can’t find my email you aren’t trying very hard) and I’ll post it up for all to see, and for your own greater glory.
### Like this:
Posted by John Armstrong | Category theory
## 6 Comments »
1. John — it’s good to see you’re introducing enriched category theory in your blog. It’s always been one of my favorite chapters in the story of category theory.
But I’m surprised that so far as I can tell, you haven’t yet mentioned in this thread that some of the most potent examples for the development of enriched category are where the monoidal category V is enriched in itself! Were you going to say something about that?
For those interested: this is where V-category theory begins to become truly “autonomous”: where V begins to play the same fundamental role in V-Cat that Set plays in ordinary category theory. The world of V-categories, or V-Cat, becomes truly self-sufficient once the fundamental constructions of ordinary category theory, e.g., functor categories, are appropriately internalized within that world. Explorations of such autonomous worlds then reveal phenomena (e.g., the importance of *weighted limits and colimits*) which yield new insights even for ordinary categories!
I’ll (mostly) resist the temptation to explain the archetypal way in which this happens, since John may want to address this himself and tie it in with other things he’s discussed. But it can be summarized in just two words: “closed categories”. I’ll add that the full autonomous development of V-category theory becomes possible (only) if V is sufficiently nice — precisely, a complete, cocomplete, symmetric monoidal closed category — and in discussions of enriched category theory it’s usually a good exercise to check for such structure. Examples include the category 2 of the entry above, Ab, of course Set… How does it work out for pSet (pointed sets)? It’s a symmetric monoidal category under cartesian product; is pSet cartesian closed? (How does one check for that?) What more can one say about this situation?
(Possibly jumping the gun, this bears on the ‘zero morphism’ conundrum posed in the entry, but for the moment I’ll leave that thread dangling, in case John is awaiting more emails on this puzzle.)
Comment by Todd Trimble | August 17, 2007 | Reply
2. I may well come back to those sorts of things. I wanted to skim enriched categories in general before moving on to abelian categories rather than just giving the definition of an $\mathbf{Ab}$-category on its own as most authors do. Then there’s homological algebra, topology, analysis, geometry…
There’s so much material out there that I’ll let some of it slide at a first pass. I’m intending to work on this thing for a good long time, so I’m comfortable waiting and coming back to subjects later.
Comment by | August 17, 2007 | Reply
3. Okay. In that case allow me to mention, for readers who might like to pursue this now: the canonical reference for enriched category theory is the late Max Kelly’s book, Basic Concepts of Enriched Category Theory. It is now available online:
http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html
Comment by Todd Trimble | August 17, 2007 | Reply
4. [...] with Zero Morphisms Todd Trimble cleared up the confusion with pointed sets and zero morphisms. The problem is that my sources were wrong to say that the [...]
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5. [...] Hom Functors As Todd Trimble pointed out, things get really nice when a category is enriched over itself. That is, the morphisms from one [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://math.stackexchange.com/questions/99850/how-can-i-solve-the-differential-equation-yy2-fx
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How can I solve the differential equation? $y'+y^{2}=f(x)$
$y'+y^{2}=f(x)$
I know how to find endless series solution via endless integral or endless derivatives , and power series solution method if we know $f(x)$. And also I know how to find general solution if we know one particular solution ($y_0$)
I am looking for exact analitic solution $y= L({f(x)})$ without knowing a particular solution, if it exists. ///$L$ defines operator such as integral,derivatives, radicals, or any defined function.
Or If it does not exist. Could you please prove why we cannot find it.
Note:This equation is related to second order differantial linear equation. If we put $y=u'/u$ This equation will turn into $u''(x)-f(x).u(x)=0$. If we find general solution of $y'+y^{2}=f(x)$, it means that $u''(x)-f(x).u(x)=0$ will be solved as well. As we know, many function such as Bessel function or Hermite polinoms and so many special functions are related to Second order linear diff equation.
Thank you for answers
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2
– Sasha Jan 17 '12 at 13:41
3
@Sasha: It is Riccati and not Ricatti. – Jon Jan 17 '12 at 13:59
2 Answers
In general case differential equations of the form : $\frac{dy}{dx}=f(x,y)$ can be rewritten as :
$M(x,y)dx+N(x,y)dy=0$
So , we have that :
$(y^2-f(x))dx +1\cdot dy=0$
Since this equation isn't exact it must be non-exact and therefore we have to find out if there exist integrating factor in terms of just one variable ($x$ or $y$) .
But , one can show that such integrating factor doesn't exist , therefore you cannot apply integrating factor method on this differential equation . So , you have to use "Riccati" method which means that particular solution has to be known .
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I don't think it is possible to show that such an integrating factor doesn't exist. – Shahab Jan 17 '12 at 15:19
@Shahab,Then please prove your assumption... – pedja Jan 17 '12 at 15:36
Oops my bad! I didnt read your comment completely. I thought you said no integrating factors will exist at all, whereas you were only referring to integrating factors in terms of only one variable. – Shahab Jan 17 '12 at 16:08
@Shahab,It always exist only if it is in terms of two variables... – pedja Jan 17 '12 at 16:09
Yeah it was a mistake on my part. – Shahab Jan 17 '12 at 16:09
Interesting. In Maple I tried $y'+y^2 = \sin(x)$, and the solution involves Mathieu functions $S, C, S', C'$.
I tried $y'+y^2=x$, and the solution involves Airy functions Ai, Bi.
I tried $y'+y^2=1/x$, and the solution involves Bessel functions $I_0, I_1, K_0, K_1$.
This is a Riccati equation. For more info, in particular how its solutions are related to solutions of a second-order linear equation, look that up.
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any second order linear diff equation ($y''(x)+p(x)y'(x)+r(x)y(x)=0$) can be transform into ($u''(x)-f(x)u(x)=0$). Thus all second order equations such as Bessel, Airy etc is directly related to my question. the equation is minimum term of all second order diff equations.I believe it is like Abel impossibilty theorm that he shew for quintic (such as $x^5+x+a=0$)cannot be solved by radicals. But I dont know how to prove that it is impossible – Mathlover Jan 17 '12 at 19:42
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http://math.stackexchange.com/questions/135042/help-me-prove-the-identity-overlinef0-frac12-pi-int-02-pi-frace?answertab=oldest
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# Help me prove the identity $\overline{f(0)} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi$
Let f be an analytic function defined in an open set containing the closed unit disk and let z in ℂ be fixed. I've simplified a more complicated expression down to this identity, and as implausible as it looks, after some numerical checking it does in fact appear to be true, although if you think you can find a counterexample be my guest:
$$\overline{f(0)} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi$$
The functions I tried were: z, z+1, and z+i, so nothing transcendental. We don't know that $\overline{f(z)}$ is analytic and we can't even push the conjugate inside the function, thus I feel like I don't have many tools at my disposal except for algebraic manipulation, and so far that hasn't gotten me anywhere.
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## 5 Answers
Here is a simple counterexample. Let $f(z) = 1$. Change variables, let $\zeta = e^{i\phi}$. Then your identity claims $$1 = \frac{1}{2\pi i} \int_\gamma \frac{d\zeta}{\zeta-z}$$ where $\gamma$ is the unit circle centered at the origin. But this integral depends on whether $z$ is in the unit circle or not. If $z$ is not in the unit circle your identity becomes $1 = 0$.
Addendum: Let's look at the integral more closely.
The claim is $\overline{f(0)} = I$, where $$\begin{eqnarray*} I &=& \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi \\ &=& \frac{1}{2\pi i} \int_\gamma \frac{d \zeta}{\zeta-z} \overline{f(\zeta)}. \end{eqnarray*}$$ We assume $z$ is not on the contour. Take the conjugate of $I$,
$$\begin{eqnarray*} \overline I &=& -\frac{1}{2\pi i} \int_\gamma \frac{\overline{d\zeta}}{\overline\zeta-\overline z} f(\zeta) \\ &=& \frac{1}{2\pi i} \int_\gamma d\zeta \, \frac{\overline\zeta}{\overline \zeta - \overline z}\frac{f(\zeta)}{\zeta} \\ &=& \frac{1}{2\pi i} \int_\gamma d\zeta \,\left[ \frac{f(\zeta)}{\zeta} - \frac{\overline z f(\zeta)}{\zeta\overline{z}-1} \right]. \end{eqnarray*}$$ where we have used $d (\overline \zeta\zeta) = \zeta d \overline \zeta + \overline \zeta d \zeta = 0$.
If $z = 0$ we find $\overline I = f(0)$. If $z\ne 0$, there are singularities at $\zeta = 0$ and $1/\overline z$. When $|z| < 1$ we don't pick up the residue at $1/\overline z$ (since this point is not in the disk) and so again, $\overline I = f(0)$. If $|z| > 1$ we get the contribution from the other singularity, and so $\overline I = f(0) - f(1/\overline z)$. Therefore, $$\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi = \begin{cases} \overline{f(0)} & |z| < 1 \\ \overline{f(0)} - \overline{f(1/\overline z)} & |z| > 1. \end{cases}$$
Going back to our counterexample for the original formula, notice for $f(z) = 1$ we find $1 = 1$ for $|z| < 1$ and for $|z|>1$ we get $0 = 1 - 1$, as we should.
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You're right, my professor may have forgotten to stipulate that z be in the unit circle, since if it is I think I came up with a proof by expanding f as a power series centered at zero. – lithium barbie doll Apr 22 '12 at 1:58
@Thoth: That is likely, the formula is not correct unless $|z|<1$. What an interesting result! – oen Apr 22 '12 at 4:28
I agree, the integral always equaling the conjugate of f at a single point is for me a very counterintuitive. – lithium barbie doll Apr 22 '12 at 5:34
In addition to the above comment by oenamen, z cannot be ON the unit circle. So I would suggest we look at the problem as if z is interior to the unit circle. I'm suspicious under this modification that it IS true. Indeed I'm staring at the reference in Reinhold Remmert's, Theory of Complex Functions which says: If f is analytic in a neighborhood of the closure of the disc B = Bs(0) (the OPEN disc of radius s centered about the origin), then If g=f conjugate, then g(0) is the integral you wrote for all z in B. The proof is based on the fact that h(w)=z*f(w)/(s^2-z*w) is holomorphic in a neighborhood of the closure of B. z* = z conjugate in the preceding. I see you wanted a proof and not just to know it's true. Well, I'll furnish the 'it's true, it's true' salute, and see if that helps for now.
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Changing variables from $\phi$ to $-\phi$ in your integral, you get $$\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{-i\phi}}{e^{-i\phi} - z}\overline{f(e^{-i\phi})}\,d\phi$$ $$= \frac{1}{2\pi}\int_0^{2\pi}\frac{1}{1 - ze^{i\phi}}\overline{f(e^{-i\phi})}\,d\phi$$ $$= \frac{1}{2\pi i}\int_0^{2\pi}\frac{1}{e^{i\phi}(1 - ze^{i\phi})}\overline{f(e^{-i\phi})}ie^{i\phi}\,d\phi$$ In contour integral form this is $$= \frac{1}{2\pi i}\int_{|\zeta| = 1}\frac{1}{\zeta(1 - z\zeta)}\overline{f(\bar{\zeta})}\,d\zeta$$
Note that $g(\zeta) = \overline{f(\bar{\zeta})}$ is analytic whenever $f$ is.
If $|z| < 1$, the integrand has a single pole, at $\zeta = 0$, with residue $g(0)$, so by the Residue Theorem the integral is $g(0) = \overline{f(0)}$ which is probably the case your teacher had in mind.
If $|z| > 1$, there's a pole at $\zeta = {1 \over z}$ with residue $-g(1/z)$ if you do the calculation, as well as the pole from before with residue $g(0)$, so by the Residue Theorem the value of the integral is actually $g(0)-g(1/z) = \overline{f(0)}-\overline{f(1/ \bar{z})}$.
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I believe your conditions on $z$ are flipped around. – oen Apr 22 '12 at 4:37
thanks, corrected it – Zarrax Apr 22 '12 at 6:40
Let $|z| < 1$ and consider:
$$\overline{f(0)} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi$$
Expand f as a power series centered at zero.
Then $\overline{f(0)} = \overline{a_0}$ and we obtain:
$$\overline{a_0} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{a_0 + a_1e^{i\phi} + a_2e^{2i\phi} + ...}\;\;d\phi$$
$$\;\;\;\;\;\;\;\;\;\; =\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\bar{a_0} + \bar{a_1}e^{-i\phi} + \bar{a_2}e^{-2i\phi} + ...\;\;d\phi$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{e^{i \phi}-z}\bar{a_0}e^{i\phi} + \bar{a_1} + \bar{a_2}e^{-i\phi} + \bar{a_3}e^{-2i\phi}+...\;\;d\phi$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{\bar{a_0}}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}d\phi + \frac{1}{2\pi}\sum_{n=0}^{\infty}\int_0^{2\pi}\frac{\overline{a_{n+1}}}{e^{i \phi}-z}e^{-ni\phi}d\phi$$
$$\;\;\;\;\;\;\;\;\;=\frac{\bar{a_0}}{2\pi i}\int_{\alpha}\frac{d\zeta}{\zeta-z}d\zeta + \frac{1}{2\pi}\sum_{n=0}^{\infty}\int_0^{2\pi}\frac{\overline{a_{n+1}}}{e^{i \phi}-z}e^{-ni\phi}d\phi$$
$$=\bar{a_0} + \frac{1}{2\pi}\sum_{n=0}^{\infty}\overline{a_{n+1}}\int_0^{2\pi}\frac{1}{e^{i \phi}-z}e^{-ni\phi}d\phi.$$
Now we need to show that for all $n\geq 0$
$$\int_0^{2\pi}\frac{1}{e^{i \phi}-z}e^{-ni\phi}d\phi = 0.$$
Making the substitution $u = e^{i\phi}$ we obtain:
$$-i\int_{\alpha}\frac{1}{u^{n+1}(u-z)}du.$$
From here one can either perform partial fraction decomposition or use the residue theorem to prove that:
$$-i\int_{\alpha}\frac{1}{u^{n+1}(u-z)}du = 0.$$
Note that $\alpha$ is the parametrization of the unit circle.
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Take complex conjugates of both sides: you're saying $$f(0) = \frac{1}{2\pi} \int_0^{2\pi} \frac{e^{-i\phi}}{e^{-i\phi}-z} f(e^{i\phi})\ d\phi$$ Now take $\zeta = e^{i\phi}$, $d\zeta = i e^{i\phi}\ d\phi$, so if $C$ is the positively oriented unit circle the right side is $$\frac{1}{2\pi i} \oint_C \frac{\zeta^{-2}}{\zeta^{-1} - z} f(\zeta)\ d\zeta = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{1-\zeta z} \frac{d\zeta}{\zeta}$$ If $|z| < 1$, $f(\zeta)/(1 - \zeta z)$ is analytic in a neighbourhood of the unit disk, so by the Cauchy integral theorem the result is $f(0)/(1 - 0 z) = f(0)$.
On the other hand, as others have remarked, if $|z| > 1$ the result is false: you'd have to take into account the residue at $\zeta = 1/z$, obtaining $f(0) - f(1/z)$.
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http://mathhelpforum.com/math-challenge-problems/22996-r-2-2-r-4-2-r-2-a.html
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Thread:
1. (r - 2)^2 + (r - 4)^2 = r^2?
How do I find the value of r?
(r - 2)^2 + (r - 4)^2 = r^2
I know what r is, coincidentally, (it's 10)
But how would I go about finding it if I didn't know?
Wasn't sure where to post this, it's not homework, I got it from a brain teaser (finding the radius of a circle which is inside a square with a 2x4 rectangle in the corner, the radius is r as seen above)
2. Originally Posted by Variant
How do I find the value of r?
(r - 2)^2 + (r - 4)^2 = r^2
I know what r is, coincidentally, (it's 10)
But how would I go about finding it if I didn't know?
Wasn't sure where to post this, it's not homework, I got it from a brain teaser (finding the radius of a circle which is inside a square with a 2x4 rectangle in the corner, the radius is r as seen above)
expand what's in the brackets and simplify to get a quadratic.
$(r - 2)^2 + (r - 4)^2 = r^2$
$\Rightarrow r^2 - 4r + 4 + r^2 - 8r + 16 = r^2$
$\Rightarrow r^2 - 12r + 20 = 0$
now continue
3. Do you know how to expand a $(a+b)^2$ ?
$(a+b)^2 = a^2 +2ab + b^2$
so $(r - 2)^2 = r^2 - 4r + 4$
and $(r - 4)^2 = r^2 - 8r + 16$
so your equation becomes $r^2 - 4r + 4 + r^2 - 8r + 16 = r^2$
then you collect like terms to get
$r^2-12r+20 = 0$
now you factor
$(r-10)(r-2) = 0$
and just you finished it off from there
edit: Jhevon beat me
4. Ahh yes. I didn't quite understand at first, how to arrive at
is what trips me up I think. I can grasp everything you just wrote, I was able to test it... but I just don't get my head around the workings of it heh.
I think I'm a lot closer to understanding it than I was before though.
Mind you, I never went much past basic math and some algebra in school, I just do these kinds of things for fun and learning.
I also think my understanding is "contaminated" by knowing the value of r.
What you guys have shown me is useful information though for learning how to do this.
5. Originally Posted by Variant
Ahh yes. I didn't quite understand at first, how to arrive at
is what trips me up I think. I can grasp everything you just wrote, I was able to test it... but I just don't get my head around the workings of it heh.
1. Initial equation
$(r - 2)^{2} + (r - 4)^{2} = r^{2}$
2. Because $a^{2}=a*a$ (when you have parentheses up against eachother as we will, it is understood that they are to be multiplied) you can rewrite the initial equation like this:
$(r - 2)(r-2) + (r - 4)(r-4) = r^2$
3. Using the technique called FOIL: first, outer, inner, last. Which basically boils down to (a+b)(c+d)=ac+ad+bc+bd.
So $(r-2)(r-2)=r*r+(-2)r+(-2)r+(-2)(-2)$ which can be simplified to $r*r-2r-2r+4$ which can again be simplified to $r^{2}-4r+4$
and using the same technique: $(r-4)(r-4)= r^{2}-8r+16$
4. So applying what we just went over:
$r^{2} -4r+4 +r^{2}-8r+16 = r^2$
5. combine like terms such as $r^{2}+r^{2}=2r^{2}$ and so on
$2r^{2} -12r+20= r^2$
6. Subtract $r^{2}$ from both sides (you must do it to both sides, because if you only did it to one side they would no longer be equal. For example, 5=5, if you subtract 1 from either side, you must subtract it from the other, so 4=4, if you do not do it to both sides, they will no longer be equal)
$r^{2} -12r+20=0$
That is how you get to this step.
edit: I can break it down a little more if you are really struggling with any given step here, let me know.
6. Oh wow, I actually got it!! Thank you!
I was thinking I was a bit dumb there for a while but I actually understand it now.
Haha, I went back to the same brain teaser, and it happened to have different values this time but I was actually able to figure it out.
7. Here is a quick way to do it. Let R = r − 2:
$\mbox{Then }(r-2)^2+(r-4)^2\ =\ r^2$
$\Rightarrow\ R^2+(R-2)^2 =\ (R+2)^2$
$\begin{array}{rcl}<br /> <br /> \Rightarrow\ R^2 &=& (R+2)^2-(R-2)^2\\\\<br /> {} &=& [(R+2)+(R-2)][(R+2)-(R-2)]\\\\<br /> {} &=& 2R\cdot4\ =\ 8R<br /> \end{array}$
$\Rightarrow\ R^2-8R\ =\ 0$
$\Rightarrow\ R(R-8)\ =\ 0$
$\Rightarrow\ R =\ 0\ \mbox{or }R\ =\ 8$
Substitute back and you’ll get r = 2 or r = 10.
8. Originally Posted by JaneBennet
Here is a quick way to do it. Let R = r − 2:
$\mbox{Then }(r-2)^2+(r-4)^2\ =\ r^2$
$\Rightarrow\ R^2+(R-2)^2 =\ (R+2)^2$
$\begin{array}{rcl}<br /> <br /> \Rightarrow\ R^2 &=& (R+2)^2-(R-2)^2\\\\<br /> {} &=& [(R+2)+(R-2)][(R+2)-(R-2)]\\\\<br /> {} &=& 2R\cdot4\ =\ 8R<br /> \end{array}$
$\Rightarrow\ R^2-8R\ =\ 0$
$\Rightarrow\ R(R-8)\ =\ 0$
$\Rightarrow\ R =\ 0\mbox{ or }R\ =\ 8$
Substitute back and you’ll get r = 2 or r = 10.
That is a cute method ^_^
9. Originally Posted by JaneBennet
Here is a quick way to do it. Let R = r − 2:
$\mbox{Then }(r-2)^2+(r-4)^2\ =\ r^2$
$\Rightarrow\ R^2+(R-2)^2 =\ (R+2)^2$
$\begin{array}{rcl}<br /> <br /> \Rightarrow\ R^2 &=& (R+2)^2-(R-2)^2\\\\<br /> {} &=& [(R+2)+(R-2)][(R+2)-(R-2)]\\\\<br /> {} &=& 2R\cdot4\ =\ 8R<br /> \end{array}$
$\Rightarrow\ R^2-8R\ =\ 0$
$\Rightarrow\ R(R-8)\ =\ 0$
$\Rightarrow\ R =\ 0\ \mbox{or }R\ =\ 8$
Substitute back and you’ll get r = 2 or r = 10.
this won't be neat once you try to solve other similar problems however.
hence this seems to be a "one trick pony" if you know what i mean.
10. Originally Posted by daddy
this seems to be a "one trick pony" if you know what i mean.
Yeah, I know what you mean, my ex-wife was the same way.
11. Originally Posted by daddy
this won't be neat once you try to solve other similar problems however.
hence this seems to be a "one trick pony" if you know what i mean.
it is a nice enough method to note. as there are a lot of problems that it would work for. the technique of substitution is used all over math, so i wouldn't file this approach under the "one trick pony" file.
in this case however, it didn't make life much simpler, so i see no harm in sticking with the straight forward approach...not that Jane was saying anything different
Originally Posted by angel.white
Yeah, I know what you mean, my ex-wife was the same way.
as in a "one trick pony"? (that's not a nice thing to call your ex-wife )...what do you mean?
12. Originally Posted by daddy
this won't be neat once you try to solve other similar problems however.
hence this seems to be a "one trick pony" if you know what i mean.
The reason for the substitution is that the equation $(r-2)^2+(r-4)^2\ =\ r^2$ is symmetrical about $r-2$. For other similar problems, you make similar substitutions involving whatever the equation is symmetrical about. For example …
$\color{white}.\quad.$ Solve $(x+123)^2+(x+456)^2=(x+789)^2$.
$\color{white}.\quad.$ $(x+123)^2+(x+456)^2=(x+789)^2$
$\Rightarrow\ x^2+246x+15129+x^2+912x+207936=x^2+1578x+622521$
$\Rightarrow\ x^2+\ldots$ ???
<screams in terror and gives up>
My method
The equation is symmetrical about $x+456$ so let $X=x+456$. Then
$\color{white}.\quad.$ $(x+123)^2+(x+456)^2=(x+789)^2$
$\Rightarrow\ (X-333)^2+X^2=(X+333)^2$
$\Rightarrow\ X^2-666X+333^2+X^2=X^2+666X+333^2$
$\Rightarrow\ X^2-1332X=0$
$\Rightarrow\ X=0$ or $X=1332$
$\Rightarrow\ x=-456$ or $x=876$
You see the difference between your method and mine? Therefore my method is not a one-trick pony! Next time think before making another similar comment.
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http://www.boredofstudies.org/wiki/Maths_Extension_2:Integration
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# Maths Extension 2:Integration
BikiCrumbs: Integration
### From Biki
Why not add to it? Don't be intimidated - we welcome all contributions!
## Integration by Substitution
Examples: For, $\int\sqrt{a^2 + x^2}\,dx$ use the substitution x = tanθ.
## Partial Fractions & Rational Functions
### Handy Partial Fractions Trick
E.g. Convert $\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ into partial fraction form.
You start off with letting $\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)} \equiv \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\,$
One way to solve the problem is to multiply both sides to get two polynomials and equate the coefficients and solve a few simultaneous equations. The other way which is not taught in HSC but is useful to know is the Residue Theorem. This is somewhat related to Remainder/Factor theorem.
To use this, you simply multiply a factor in the denominator and then substitue the "zero" into the fraction, so:
• To find $A\,$, substitue $x=1\,$ in $(x-1)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ so you get
• $A = \frac{15(1)^2-56(1)+47}{((1)-2)((1)-3)}=\frac{6}{2}=3\,$
• To find $B\,$, substitue $x=2\,$ in $(x-2)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ so you get
• $B = \frac{15(2)^2-56(2)+47}{((2)-1)((2)-3)}=\frac{-5}{-1}=5\,$
• To find $C\,$, substitue $x=3\,$ in $(x-3)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ so you get
• $C = \frac{15(3)^2-56(3)+47}{((3)-1)((3)-2)}=\frac{14}{2}=7\,$
Hence $\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)} \equiv \frac{3}{x-1}+\frac{5}{x-2}+\frac{7}{x-3} \,$
This method only works if the factor you are working with looks like $x-a\,$ and not $x^n+a\,$ or $(x-a)^n\,$.
## Integration by Parts (Reverse Product Rule)
Integration by parts allows the student to perform integration of the form $\int f(x)g(x)dx$ with relative ease.
Students of Mathematics Extension 2 would already be familiar with the rule for differentiating the product of two functions: $\frac{d}{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)$. Integration by parts uses this fact to construct a rule applicable to integration, as follows:
Suppose G'(x) = g(x); that is, suppose the primitive of g is G. Thus, by the product rule and recasting:
$\frac{d}{dx}f(x)G(x)= f'(x)G(x) + f(x)g(x)$
$f(x)g(x) = \frac{d}{dx}f(x)G(x) - f'(x)G(x)$
Integrating both sides gives:
$\int f(x)g(x)dx$ $=\int \frac{d}{dx} f(x)G(x) - \int f(x)G(x)dx$
$=f(x)G(x) - \int f'(x)G(x)dx$
which is the rule of integration by parts.
## Reduction Formulae
For example:
Find $\int sin^nx\, dx$
and hence find $\int_{0}^{\frac{\Pi}{4}} sin^9x\, dx$
Solution:
Let
$\int sin^nx=I_n \,$
And let
$sin^nx = sin^{n-1}*sinx\,$
Integrating by parts
$u = sin^nx \,$
$u'= (n-1)(sin^{n-2}x)(cosx) \,$
$v'=sinx\,$
$v = -cosx\,$
NOTE: To evaluate $u'\,$, use the chain rule.
$I_n = uv-\int u'v\, dx$
$I_n = -sin^nxcosx-\int (n-1)(sin^{n-2}x)(cosx)(-cosx)\, dx$
Take a factor of (n − 2) out of the integral.
$I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x)(cos^2x)\, dx$
Using $sin^2x+cos^2x=1\,$ substitute $cos^2x=(1-sin^2x)\,$
$I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x)(1-sin^2x)\, dx$
$\int (sin^{n-2}x)(1-sin^2x)\, dx$ $= \int (sin^{n-2}x) - \int (sin^nx)\,$ (by expanding and separating)
$I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x) - (n-1)\int (sin^nx)\,$
Add $(n-1)\int sin^nx\,$ to both sides.
For the Left side you have:
$\int (sin^nx)+n\int (sin^nx)-\int (sin^nx)\,$
$= n\int (sin^nx)\,$
$n\int (sin^nx) = -sin^nxcosx+(n-1)\int (sin^{n-2}x)\,dx$
Divide by n
$\int (sin^nx) = \frac{-sin^nxcosx}{n}+\frac{n-1}{n}\int (sin^{n-2}x)\,dx$
This is our general solution for $\int sin^nx \,$
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http://math.stackexchange.com/questions/124047/is-the-clustering-of-prime-powers-merely-coincidental
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# Is the clustering of prime powers merely coincidental?
$2^3$ and $3^2$ are close together; $11^2$, $5^3$, and $2^7$ (121, 125, and 128) are close together; $3^5$, $2^8$, and maybe $17^2$ (243, 256, and 289) are close together. $7^3$ is close to $19^2$ (343 and 361). $3^7$ is very nearly $13^3$ (2187 and 2197), which is very nearly $47^2$ (2209). $19^4$ is close to $2^{17}$ (130,321 and 131,072). Further examples are easy to find.
One might expect coincidences to get farther apart and rarer as the numbers get larger, but that doesn't seem to happen. For example, $13^{11}$ and $23^9$ differ by about one part in 200 (1,792,160,394,037 and 1,801,152,661,463).
Is this all just the law of small numbers at work? How many such coincidences would one naïvely expect? Is there any evidence that the number of such coincidences is, or is not, more than one would naïvely expect? Is anything known about the distribution of prime powers?
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It occurs to me that one should consider numbers of the form $n \log p$ instead of $p^n$. Each sequence of $n \log p$ for fixed $p$ is arithmetic, but with a different common difference than the other sequences. $$\qquad$$ As we go farther out, we are merging more and more of these sequences, so perhaps it's not surprising at all that the values get closer than any given $\epsilon$ more and more frequently.$$\qquad$$ I have more notes about this but they don't fix in the box. I'd still like to hear what other people have to say. – MJD Mar 24 '12 at 21:26
3
Are you familiar with the abc conjecture? This would put some limits on how close prime powers can be. – Gerry Myerson Mar 24 '12 at 21:44
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– Jyrki Lahtonen Mar 24 '12 at 21:50
@GerryMyerson: I was not familiar with the abc conjecture, but I read up on it last night. Do I understand correctly that your idea is that $a$ and $c$ will be prime powers and then the abc conjecture, if true, would bound how often rad(b) would be small? – MJD Mar 28 '12 at 15:36
Yes, that's the idea. – Gerry Myerson Mar 29 '12 at 5:49
## 1 Answer
For primes $p$ and $q$ the ratio $r=\log(p)/\log(q)$ is an irrational number and by the Equidistribution theorem the sequence $\{r,2r,3r,4r,\ldots\}$ is asymptotically equidistributed modulo 1.
Specifically, for large $N$ we would expect that $2\epsilon N$ elements of $\{i\cdot r\}_{i=1}^N$ are equivalent mod 1 to a number in the intervals $[0,\epsilon]\cap[1-\epsilon,1)$. This implies that $p^i$ is within a factor of $q^\epsilon$ of a power of $q$.
So if we look at powers of $p$ up to $N$ and want one to be within about 1 part in 200 of a power of $q$, approximately we want $\left|\log\left(p^a/q^b\right)\right|\le 0.005$, then we would expect to find $2N\log(1.005)/\log(q)$ close pairs.
Below are some charts showing this estimate and the actual number of pairs of exponents that give powers within 0.005 for pairs of primes $2\le p<q \le 29$. The x-axis enumerates the prime pairs, starting at $(2,3)$ and ending at $(23,29)$. The 17th entry showing a count of 2 in the first chart is $(3,17)$.
Note that the above argument does not rely on $p,q$ being primes, only that $\log(p)/\log(q)$ is irrational. Here are tables for $N=100$ and $N=10000$ for a few sets of primes as well as $(2,\pi)$ and $(\zeta(3),\mathrm{e})$.
$$\begin{array}{|c|c|ccc|} p & q & Best & |\log| & \#\{|\log|\le 0.005\} & Expected \\ \hline 2 & 3 & 2^{84} \sim 3^{53} & 0.0021 & 1 & 0.9 \\ 2 & 5 & 2^{65} \sim 5^{28} & 0.0097 & 0 & 0.6 \\ 5 & 13 & 5^{51} \sim 13^{32} & 0.0030 & 1 & 0.4 \\ 3 & 17 & 3^{49} \sim 17^{19} & 0.0009 & 2 & 0.4 \\ 13 & 17 & 13^{95} \sim 17^{86} & 0.0138 & 0 & 0.4 \\ 11 & 23 & 11^{17} \sim 23^{13} & 0.0028 & 1 & 0.3 \\ 17 & 29 & 17^{82} \sim 29^{69} & 0.0199 & 0 & 0.3 \\ 1229 & 1381 & 1229^{62} \sim 1381^{61} & 0.0009 & 1 & 0.1 \\ 2 & \pi & 2^{71} \sim \pi^{43} & 0.0099 & 0 & 0.9 \\ \zeta(3) & \mathrm{e} & \zeta(3)^{38} \sim \mathrm{e}^{7} & 0.0067 & 0 & 1.0 \\ \hline \end{array}$$
$$\begin{array}{|c|c|ccc|} p & q & Best & |\log| & \#\{|\log|\le 0.005\} & Expected \\ \hline 2 & 3 & 2^{1054} \sim 3^{665} & 0.00004 & 90 & 90.8 \\ 2 & 5 & 2^{9297} \sim 5^{4004} & 0.00006 & 62 & 62.0 \\ 5 & 13 & 5^{9551} \sim 13^{5993} & 0.000002 & 38 & 38.9 \\ 3 & 17 & 3^{5965} \sim 17^{2313} & 0.00016 & 37 & 35.2 \\ 13 & 17 & 13^{1637} \sim 17^{1482} & 0.00008 & 34 & 35.2 \\ 11 & 23 & 11^{4489} \sim 23^{3433} & 0.00024 & 30 & 31.8 \\ 17 & 29 & 17^{2875} \sim 29^{2419} & 0.00025 & 28 & 29.6 \\ 1229 & 1381 & 1229^{7813} \sim 1381^{7687} & 0.00012 & 16 & 13.8 \\ 2 & \pi & 2^{9217} \sim \pi^{5581} & 0.00007 & 86 & 87.1 \\ \zeta(3) & \mathrm{e} & \zeta(3)^{1641} \sim \mathrm{e}^{302} & 0.00008 & 98 & 99.8 \\ \hline \end{array}$$
It doesn't seem that the number of close pairs is more than expected. For $N=100$ and a 0.005 cutoff the average count is slightly less than we might expect from the asymptotics, but by $N=10000$ the observed match the model quite closely.
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What a beautiful answer! – Bruno Mar 29 '12 at 4:41
Thanks!......... – Zander Mar 29 '12 at 13:03
I will probably accept this soon, but I would like to poke around with it a little more. I did not want you to think I had not seen and appreciated your answer, or that I had forgotten about it. Thanks. – MJD Apr 13 '12 at 14:47
Thanks again for the effort you put into this answer. I am very grateful. – MJD May 11 '12 at 1:05
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http://physics.stackexchange.com/questions/8466/sun-symmetry-and-its-representations
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# SU(N) symmetry and its representations
If a Lagrangian containing an N-multiplet of fields is invariant under global $\mathbf{SU}(N)$ transformations, does that necessarily imply it is invariant under $\mathbf{SU}(N-1)$, $\mathbf{SU}(N-2)$,... all the way down to $\mathbf{U}(1)$ as well, where the representations in each case are $N$x$N$ matrices?
Also, given a symmetry group $\mathbf{SU}(N)$, do representations involving $M$x$M$ matrices exist for all $M \geq N$?
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1
The latter question about representations is something that would be better asked on math.SE. I'd suggest that you edit it out and repost it over there. – David Zaslavsky♦ Apr 11 '11 at 20:38
I believe you are asking: Is an (N+M)-dimensional representation of an element of SU(N) also an element of SU(N+M)? – Matt Apr 12 '11 at 4:45
To answer the last part of the question(v1) with the understanding that OP is only interested in irreps: No, one counterexamples would be $su(3)$, which does not have a $4$-dimensional irrep, corresponding to $M=4\geq 3=N$. – Qmechanic♦ Feb 19 '12 at 14:09
## 1 Answer
Well, if something is invariant under the action of some group it means that every element leaves it invariant. In particular, every element that belongs to some subgroup. Also, you can get an obvious representation of the subgroup by restriction. If $\rho: G \to {\rm Aut}(V)$ is a representation then so is $\rho_{H}: H \to {\rm Aut}(V)$ for $H < G$. Naturally, this need not be irreducible anymore (obvious example being representation of a trivial subgroup which is reducible to $N$ components).
Well, they surely do because one can take $T\oplus \dots \oplus T$ where $T$ is trivial one dimensional representation. Let's pretend that you meant irreducible representations so that this question is actually non-trivial. Groups of $A_l$ type (with $l = N-1$) are of rank $l$. This means that finite-dimensional representations (as is the case for the irreducible representations of compact $\mathbf {SU}$ forms of these groups) are indexed by $l$ non-negative integers $\{\lambda_i\}_{i=1}^l$. E.g. for $l=1$ (${\mathbf {SU}(2)}$) this integer is connected to spin by $\lambda_1 = 2s$ (and dimension of these representations is $\lambda_1 +1 = 2s+1$). So we are interested in determining dimensions of the irreducible representations indexed by $\lambda$. This is not so straight-forward; in principle one can determine these dimensions by looking at the weight space of the given representation and counting the number of ways one can arrive at any given weight starting from highest weight and applying annihilation operators.
Instead, one can exploit the fact that $\mathbf {SL}$ is basically the same thing as $\mathbf {GL}$ and use Schur-Weyl duality between $\mathbf {GL}(N,\mathbb C)$ and $\mathfrak{S}_k$ (i.e. the symmetry group of $k$-element set) which states that the representation of the above groups on $(\mathbf C^n)^{\otimes k}$ (with obvious actions) decomposes as sum over partitions $\lambda \in Par(k,N)$ (e.g. $(3,2,1) \in Par(6,3)$) of products of irreducible representations $F^{\lambda} \otimes G^{\lambda}$ (with $F$ resp. $G$ being a representation of $\mathbf {GL}(N,\mathbb C)$, resp. $\mathfrak{S}_k$). The dimensions of these representations can then be computed by fancy combinatorial formulas derived from Young tableaux (which enter due to partitions, of course). Basically one writes down some numbers related to $n$ and number of rows under and columns to the right into each box of the tableaux and multiplies all those numbers together. As for whether every integer greater or equal to $N$ can be obtained in this way, I don't know. I suppose by shear amount of these tableaux this should be true but I'll come back to check it later.
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I don't think that, given some $N$, there are $M$-dimensional irreps for $SU(N)$ for all $M > 1$. For $SU(2)$ it's true, but even for $SU(3)$ I don't believe there are irreps of dimension 2, 4 and 5 (for example). I haven't really thought about it, but if these exist, do tell! – Vibert Dec 20 '12 at 19:06
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http://mathhelpforum.com/calculus/168759-volume-involving-triple-integral.html
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# Thread:
1. ## Volume involving a triple integral
Find the volume determined by z less than or equal to 6-x^2-y^2 and z greater than or equal to sqrt(x^2+y^2).
I'm not sure how to find the limits of integration for this question. I set the two inequalities to each other and found that x^2+y^2=9 and x^2+y^2=4, but am a little stumped on where to continue from there. I'm thinking at some point I will need to convert to cylindrical coordinates, but am still not sure how to go about it. Any help would be appreciated.
2. In cylindrical the equations are $z=6-r^2$ and $z=r$. These intersect in the circle $r=2$.
So in cylindrical, we have $0\leq \theta \leq 2\pi , 0\leq r\leq 2$ and $r\leq z\leq 6-r^2$.
3. Thanks so much! I ended up getting 32pi/3, which I hope is right!
4. Yep - that's what I got.
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http://crypto.stackexchange.com/questions/3958/besides-key-and-ciphertext-sizes-what-are-other-advantages-of-elliptic-curve-ver?answertab=votes
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# Besides key and ciphertext sizes what are other advantages of elliptic curve versions of various protocols?
There are elliptic curve variants of Diffie-Hellman, ElGamal, DSA and possibly other protocols/algorithms. I know that these elliptic curve variants have smaller key and ciphertext sizes which will make communications (or storage) more efficient.
Are there other advantages to the elliptic curve variants of these standard protocols/algorithms?
Specifically I am interested in speed and power consumption, but other potential advantages like resistance to side-channel attacks are also of interest.
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## 2 Answers
Computations on elliptic curves are more efficient. Roughly speaking, when the base field has size $n$ (for DH/ElGamal/DSA, the size in bits of the modulus $p$; for elliptic curves, the size of the field for point coordinates) and a "security level" $t$ (e.g. $t = 80$ for "80-bit security" as can be expected when using a 160-bit subgroup and a 160-bit hash function), the computational cost of a private key operation (decryption, signature generation) is roughly $O(n^2t)$, both for the modular and the elliptic curve variants. The big-O notation hides a constant, which is about $10$: for the same $n$ and $t$, the elliptic curve variant will be 10 times slower than the older modular algorithm.
However, elliptic curves tolerate much lower $n$ for a given security level. Basically, for $t = 112$, we need $n = 2048$ for the modular algorithms, but $n = 224$ is sufficient for elliptic curves (these comparisons are always a bit subjective, see this site for details). Report the values in the formulas: even with the factor 10, the elliptic curve variant will be more than 8 times faster than the modular algorithm of comparable strength.
Another point is binary fields. When the implementation platform is a dedicated ASIC/FPGA, or even when the platform is a small CPU which does not offer a very optimized multiplication opcode (e.g. the ARM Cortex M0), then curves on binary fields give a huge implementation boost, especially Koblitz curves. It also has benefits for memory-constrained systems, see for instance this presentation. The "modular" algorithms cannot really benefit from binary field computations (it can be defined, but discrete logarithm on binary fields is easier than modulo a random prime, so a binary field variant of modular DH or ElGamal would require a $n$ twice larger, hence a four-fold slowdown).
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Elliptic Curve Cryptography (ECC) is not known to be specifically more resistant to side channel attacks (of course the next question is more resistant than what).
• This paper reviews power analysis side-channel attacks against ECC and countermeasures.
• Given that ECC uses multiplication and many common implementations of the MUL instruction run in time dependent on input (x86 on intel fixed this on newer chips), there are probably some room for timing side channels on no-intel chips.
• Or double ADD timing attacks:
"Since point doubling (P + P) requires slightly fewer processor arithmetic operations than arbitrary point addition (P + Q, P != Q) due to the ability to reduce the group law for P + P, the attacker can distinguish between these two cases, and the key can be reconstructed. This was one very concrete, simple example of how practical it is to attack an FPGA implementation with nothing more than an oscilloscope, although the countermeasures are easily implemented." - Pertinent Side Channel Attacks on Elliptic Curve Cryptographic Systems
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http://www.reference.com/browse/Exponential_utility
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Definitions
# Exponential utility
In economics exponential utility refers to a specific form of the utility function, used in many contexts because of its convenience when uncertainty is present. Formally, exponential utility is given by:
$u\left(c\right)=-e^\left\{-a c\right\}$,
where $c$ is consumption and $a$ is a constant.
Exponential utility implies constant absolute risk aversion, with coefficient of absolute risk aversion equal to
$frac\left\{-u\text{'}\text{'}\left(c\right)\right\}\left\{u\text{'}\left(c\right)\right\}=a.$
Though isoelastic utility, exhibiting constant relative risk aversion, is considered more plausible (as are other utility functions exhibiting decreasing absolute risk aversion), exponential utility is particularly convenient for many calculations. Specifically, under exponential utility, expected utility is given by:
$E\left(u\left(c\right)\right)=E\left(-e^\left\{-a \left(c+epsilon\right)\right\}\right),$
where E is the expectation operator. With normally distributed noise, ie,
$epsilon sim N\left(mu, sigma^2\right),!$
E(u(c)) can be calculated easily using the fact that
$E\left(e^\left\{-a epsilon\right\}\right)=e^\left\{-a mu + frac\left\{a^2\right\}\left\{2\right\}sigma^2\right\}.$
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http://mathoverflow.net/questions/28967/a-characterization-of-convexity/29004
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## A characterization of convexity
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
While doing some research on polytopes I came to the following question. Maybe it's already somewhere but anyway I'll post it here.
Let $X\subset \mathbb{R}^3$ be such that, for every plane $P$, $P\cap X$ is simply connected. Is $X$ convex?
I'm not sure, but I think maybe it's necesary to assume some well behaving like local simple connectedness. Anyway I think this is true with the apropiate asumptions. I would not be surprised if it was true just as stated.
Probably this is true even in greater dimensions.
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1
By simply connected, you implicitely mean connected and simply connected, right? – Benoît Kloeckner Jun 21 2010 at 16:40
For dimensions $n$ greater than 3, are you assuming that you intersect planes with $X$ or that you intersect $n-1$-planes with $X$? – supercooldave Jun 21 2010 at 16:40
n-1 planes and yes, connected and simply connected – Cristos A. Ruiz Jun 21 2010 at 22:41
## 5 Answers
How about some tomography? This should work if $X$ is open. Assume $X\subset \mathbb R^3$ is nonempty and for every plane $H$ the intersection $H\cap X$ is either contractible or empty. (Note that an open subset of the plane is contractible if it is ((nonempty,) connected, and) simply connected.)
Claim 1: $X$ is contractible.
Proof: Consider the space of all pairs $(x,H)$, $x\in X$ and $H$ a plane containing $x$.
Fiber this by $(x,H)\mapsto x$. It's a trivial bundle over $X$ with fiber $P^2$.
Fiber the same thing by $(x,H)\mapsto H$. The nonempty fibers are contractible, so the domain is homotopy equivalent to the image, which in turn fibers over $P^2$ with contractible fibers.
So $X\times P^2$ has the same homotopy type as $P^2$, and upon further inspection $X$ is contractible.
Claim 2: $X$ is convex.
Proof: Let $L$ be any line whose intersection with $X$ is nonempty, and play the same game again with pairs $(x,H)$, but now the plane $H$ is constrained to contain $L$. Call the space of all such pairs $Y$. On the one hand, $Y$ is equivalent to the circle $P^1$ by the same kind of argument as before. On the other hand, $Y$ is the blowup of the $3$-manifold $X$ along the $1$-manifold $X\cap L$. The complement $Y'$ of $(X\cap L)\times P^1$ in $Y$ is the same as the complement of $X\cap L$ in $X$, so it is connected. Therefore all of the group $H_1(Y,Y')$ comes from $H_1(Y)$. But the latter group is of rank $1$ while the former is isomorphic to $H_0( (X\cap L)\times P^1)$. (I am using mod $2$ coefficients.) It follows that $X\cap L$ is connected.
Maybe this generalizes to $\mathbb R^n$. The hypothesis I am thinking of is that every nonempty hyperplane section is contractible, or equivalently $(n-2)$-connected. I don't believe that every hyperplane section simply connected is enough if $n>3$
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I don't think that Claim 2 is true if $X$ is the interior of my modification of Diego Matessi's example: $X=\{ (x,y,z)\in\mathbb{R}^2\times\mathbb{R}_+: x<0 \text{ or } x^2<y^2+z^2\}.$ – Victor Protsak Jun 21 2010 at 22:34
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Your $X$ intersects the plane $x=z+1$ in a disconnected set. – Tom Goodwillie Jun 21 2010 at 23:51
Yes, you are right, thank you! The vertex of the parabola is below the $xy$-plane, so there are two pieces. Once I drew it, it became obvious. – Victor Protsak Jun 22 2010 at 0:38
I'm chosing this one as the correct answer. There was also Matessi's correct counterexample and Pak's reference (by the way, that book looks really good). But I think there's more merit in this one, and it looks correct to me, also I don't know if the case of $X$ open was known, and the method used to solve it is very appealing to me. Thanks very much for your responses! – Cristos A. Ruiz Jun 24 2010 at 18:28
Thanks. I don't swear that it's correct, though. – Tom Goodwillie Jun 27 2010 at 1:13
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
A version of this is well known:
Let $X\subset \Bbb R^3$ be a compact set and suppose every intersection of $X$ by a plane is contractible. Then $X$ is convex.
This is due to Schreier (1933) in $\Bbb R^3$, and Aumann (1936) generalized this to higher dimensions. See this and other related results in Ju.D. Burago and V.A. Zalgaller, Sufficient criteria of convexity, J. Math. Sci. 10 (1978). 395–435. Incidentally, this a (hard) exercise in my book (Exc. 1.25).
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In coordinates $(x,y,z)$, consider { $z \geq 0$} and remove from it the half line { $(t,0,0) | t > 0$}. You get something non convex, but it seems to me that it also satisfies your property. Maybe you need some boundedness?
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Nice example, that can be made closed by removing a semi-infinite open cylinder instead of a half-line. – Benoît Kloeckner Jun 21 2010 at 16:56
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@Benoit: Would it not then be possible to choose $P$ slanting so that it clips a bit of the base of the cylinder near the origin, and so creating a hole in $P \cap X$? Maybe I misunderstand what you intend. – Joseph O'Rourke Jun 21 2010 at 18:49
Joseph: That's right! It can be fixed by removing a half-cone $\{ (x,y,z): x>0, x^2>y^2+z^2\}$ instead. Then $P\cap X$ is $P_+\setminus Q$, where $P_+$ is a half-plane and $Q$ is empty, a point, a wedge, a region above a parabola or hyperbola, or a an elliptical segment with the chord along the boundary line of $P_+.$ – Victor Protsak Jun 21 2010 at 22:15
As Tom Goodwillie pointed out, the intersection with $x=z+1$ is disconnected (it is the region above a parabola whose vertex is below the boundary of the half-plane), so this doesn't work, either. – Victor Protsak Jun 22 2010 at 0:42
Unboundedness is not the issue. Diego Matessi's example can be modified by intersecting with max(|x|,|y|,z)<1. – Tom Goodwillie Jun 22 2010 at 2:04
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Edit Corrected.
Here is an elementary statement with a similar flavor (this does not answer the question).
Let $M$ be a compact manifold and $\mathcal{F}$ a family of functions $f: M\to \mathbb{R}^{n_f}$ with the following properties: a. $\mathcal{F}$ is closed under linear projections b. The pre-image $f^{-1}(c)\subset M$ under $f\in\mathcal{F}$ of any point $c\in \mathbb{R}^{n_f}$ is either empty or connected. Then the image $f(M)$ of $M$ under any $f\in\mathcal{F}$ is convex.
Using Morse theory, Atiyah proved that these hypotheses hold for the moment maps of Hamiltonian torus actions on a compact symplectic manifold $M$1 and concluded that the image $\mu(M)$ is a convex polytope whose vertices are the images of the fixed points.
Atiyah, Convexity and commuting Hamiltonians, Bull. London Math. Soc. 14 (1982), no. 1, 1–15
Footnote
1 More precisely, families of commuting Hamiltonians which generate a compact subgroup in $Diff(M).$
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There is something wrong: the continuous image of a connected set is connected, so that the right-hand side is satisfied for all connected $K$. – Benoît Kloeckner Jun 22 2010 at 12:26
Hmm... you are right, I need to check the paper. It's possible that the condition is really in terms of intersections with hyperplanes. – Victor Protsak Jun 22 2010 at 13:18
It is not true for non locally-connected sets. Take a path $\gamma$ in $\mathbb{R}^2$ with one end that accumulates to a subsegment of $\gamma$ (in a $\sin(1/x)$ way). Take the product of $\gamma$ with $\mathbb{R}$, and add the interior of $\gamma$ times `$\{0\}$`.
If you want it to be compact, take `$\gamma\times\{0\}$` union with a cone based on $\gamma$.
Note that $\gamma$ is also a weird example of a compact set of the plane that is pathwise connected, simply connected, and whose complement has two connected components.
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Re "weird example of a compact set": what you mean is probably a "quasicircle" (Hatcher, Alg Top, Sec 1.3, Ex 7), because the complement of the closure of the graph of $\sin(1/x), x>0$ is clearly contractible. – Victor Protsak Jun 21 2010 at 23:17
Yes, that is what I tried to describe. – Benoît Kloeckner Jun 22 2010 at 12:19
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http://physics.stackexchange.com/questions/1873/the-concepts-of-path-integral-in-quantitative-finance
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# The Concepts of Path Integral in Quantitative Finance
I realize that path integral techniques can be applied to quantitative finance such as option value calculation. But I don't quite understand how this is done.
Is it possible to explain this to me in qualitative and quantitative terms?
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Although some physicists are interested in quantitative finance, this question is off-topic here. Unfortunately I can't point you to the appropriate forum off the top of my head, but I'm sure quantitative finance forums exist if you poke around. – Mark Eichenlaub Dec 13 '10 at 9:30
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@Mark, I do not think so; I think econophysics questions should be allowed here, much like mathematical physics, or physics questions with engineering bent are allowed here. – Graviton Dec 13 '10 at 9:36
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– user346 Dec 13 '10 at 9:51
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I voted to reopen. (I couldn't figure out how to redact my vote to close.) I don't have a strong personal stake in this question - my initial vote to close was just my immediate reaction because I hadn't seen such questions here before and I didn't know about any significant ties to physics. It now seems that a majority of users think the question is appropriate, and I'm happy to go along with the majority. – Mark Eichenlaub Dec 14 '10 at 4:51
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@marek I don't know about string theory but the complete works of shakespeare should be mandatory reading for all experimentalists :-) Theorists are just born with it ! – user346 Dec 14 '10 at 20:23
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## 1 Answer
The fundamental equation which serves as the basis for the path-integral formulation of finance and many physical problems is the Chapman-Kolmogorov equation.
$$p(X_f|X_i)=\int p(X_f|X_k)p(X_k|X_i) dX_k$$
This is analogous to the following equation for amplitudes in quantum mechanics
$$\langle X_f|X_i \rangle=\int \langle X_f|X_k\rangle\langle X_k|X_i\rangle dX_k$$
That's right, it's the same form, but the interpretation of the basic entities changes. In the former, they are probability densities and thus real and positive, in the latter they are probability amplitudes and thus complex.
The class of physical problems that can be tackled with the first type of equation are called Markov processes, their characteristic is that the state of the system depends only on its previous state. Despite its seeming limitedness, this comprises many phenomena since any process with a long but finite memory can be mapped onto a Markov process provided the state space is enlarged appropriately. On the other hand, the second equation is pretty natural and general in quantum mechanics. It is basically stating that the unity operator can always be decomposed into a, possibly overcomplete, sum of pure states
$$\mathbb{I}=\int |X_k\rangle\langle X_k| dX_k \; .$$
Now, constructing a path integral is done by slashing up the path from $X_i$ to $X_k$ into ever smaller components. Let's suppose that the endpoints are fixed, then we might assume that to go from one endpoint to another, the system has to go through paths $(X_i,X_1(t_1),X_2(t_2),\ldots,X_n(t_n),X_f)$. This leads to the following integral
$$p(X_f|X_i)=\int\cdots\int \prod_{k=0}^n p(X_{k+1}(t_{k+1})|X_k(t_k)) \prod_{k=1}^n dX_k(t_k)$$
where I put $X_0(t_0)=X_i$ and $X_{n+1}(t_{n+1})=X_f$. The tricky part is now to see if the limit can be defined meaningfully. This can be very problematic, especially in the quantum case. Ironically, the cases that are used for finance and statistical mechanics are often much more well-behaved. This is again related to one integral being over complex numbers and the other over real numbers, but it's not the only reason. Up till now, I have not been specific about the kind of system I want to study, this will play an important role as well.
So, let's take an option which is a financial security of which the price is dependent on the price of the underlying stock and time. So we can write $O(X,t)$ for the price of the option and we'll assume the underlying stock follows a geometric brownian motion:
$$\frac{dX}{X}=\mu dt + \sigma dW$$
where $W$ represents a Wiener process with increments $dW$ having mean zero and variance $dt$. Also assume that the pay-off of the option at the expiration time $T$ is:
$$O(X_T,T)=F(X_T)$$
with $F$ a given function of the terminal stock price.
Then, Fisher Black and Myron Scholes have shown that the option, under the 'no arbitrage' assumption, satisfies the following PDE
$$\frac{\partial O}{\partial t} + \frac{1}{2}\sigma^2X^2\frac{\partial^2 O}{\partial X^2} + r X \frac{\partial O}{\partial X} - rO = 0$$
in which $r$ is the risk free interest rate. If instead of the geometric brownian motion variable $X$, I reformulate this into $x=\ln X$ which is an arithmetic brownian motion variable, I can reformulate the equation as:
$$\frac{\partial O}{\partial t} + \frac{1}{2}\sigma^2\frac{\partial^2 O}{\partial x^2} + (r-\frac{\sigma^2}{2}) \frac{\partial O}{\partial x} - rO = 0$$
This is nothing else but a special case of the PDE's that can be solved by using the Feynman-Kac formula, which includes also the Fokker-Planck equation and the Smoluchowski equation, both related to the description of diffusion processes in physics. In the diffusion problem, O is to be interpreted as a distribution of velocities of the particle (Fokker-Planck) or of the positions of the particle (Smoluchowski). That's how we relate to what I introduced above. Also note that the Schrödinger equation in quantum mechanics is very similar in form, except you'll get complex coefficients.
The Feynman-Kac formula tells us that the solution to the PDE is:
$$O(X,t) = e^{-r(T-t)}\mathbb{E}\left[ F(X_T)|X(t)=X \right]$$
It is this expectation value that will now be represented as a pathintegral:
$$O(X,t) = e^{-r(T-t)}\int_{-\infty}^{+\infty}\left(\int_{x(t)=x}^{x(T)=x_T} F(e^{x_T}) e^{A_{BS}(x(t'))} \mathcal{D}x(t')\right) dx_T$$
where
$$A_{BS}(x(t'))=\int_t^{T} \frac{1}{2\sigma^2}\left(\frac{dx(t')}{dt'}-\mu\right)^2$$
is the action functional.
The reason this path integral can be built is the same explained before, here it is possible to split the conditional expectation ever further in smaller intervals:
$$\begin{array}{rcl}\mathbb{E}\left[ F(e^{x_T})|x(t)=x \right] & = & \int_{-\infty}^{+\infty} F(e^{x_T}) p(x_T|x(t)=x) dx_T \\ & = & \int_{-\infty}^{+\infty} F(e^{x_T}) \int_{\tilde{x}(t)=x}^{\tilde{x}(T)=x_T} p(x_T|\tilde{x}(\tilde{t})) p(\tilde{x}(\tilde{t})|x(t)=x) d\tilde{x}(\tilde{t}) dx_T \end{array}$$
Each of the conditional probabilities satisfying the PDE for the arithmetic brownian motion as noticed above.
I'll stop here for now, but I refer to the following article for further details.
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This doesn't talk about finance at all hardly. – Noldorin Dec 13 '10 at 20:52
It doesn't really talk about physics, either. – Sklivvz♦ Dec 13 '10 at 22:13
Excellent answer @raskolnikov! I see the peanut gallery is pretty crowded today. – user346 Dec 13 '10 at 23:29
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Seems like the villagers have taken over. I vote to reopen. – user346 Dec 14 '10 at 0:47
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I voted to reopen. – Robert Smith Dec 14 '10 at 2:18
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http://math.stackexchange.com/questions/218626/how-many-subgroups-can-we-make-of-group-whose-order-is-a-multiple-of-prime
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# How many subgroups can we make of group whose order is a multiple of prime
If a group has $n$ elements of order $p$ ($p$ is prime) how many subgroups of order $p$ do we get? If the group was cyclic than only one unique subgroup of order $p$, but what if the group is non-cyclic?
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## 2 Answers
Turn the question around: if $G$ has $m$ subgroups of order $p$, how many elements of order $p$ does it have?
If $x$ has order $p$, then so have $x^2,x^3,\dots,x^{p-1}$, and each of them generates the same subgroup of $G$, namely, $\langle x\rangle$. Thus, if $G$ has $m$ subgroups of order $p$, each of them must contain $p-1$ elements of order $p$, and since $p$ is prime, any two of these subgroups have trivial intersection. Thus, $G$ has $m(p-1)$ elements of order $p$.
So if $G$ has $n$ elements of order $p$, what is $m$ (in terms of $p$ and $n$)?
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thanks a lot for the help. – d13 Oct 22 '12 at 11:08
Is the group itself of order p or are we discussing the order of the elements of this group?
Because if the group itself is of order p (i.e. size p) then there can only be two "subgroups." We can have a subgroup of size 1 (the identity group, {e}) and a subgroup of size p which is the same size as the original group.
This arises from Lagrange's Theorem:
Let G be a finite group and H a subgroup of G. Then the size of the subgroup divides the size of the group.
I take it that G can only have subgroups that divide it.
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yes i understand what u say is right but the order of the group is not given, if it was, obviously i wouldn't ask the question here :) – d13 Oct 23 '12 at 6:33
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http://math.stackexchange.com/questions/116236/mean-value-property-with-fixed-radius
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# Mean value property with fixed radius
I will focus on the real line. Let $f$ be a smooth function on $\mathbb{R}$, if $\forall x\in\mathbb{R}, r>0$, $$\frac{f(x-r)+f(x+r)}{2}=f(x),$$ we say that f has the spherical mean value property (MVP); if instead $$\frac{1}{2r}\int_{x-r}^{x+r}f(t)dt=f(x),$$ we say f has the ball MVP. It is easy to see that spherical MVP implies ball MVP.
It is well known, and not hard to prove (though it is much harder in higher dimensions), that ball MVP implies that $f$ is harmonic, i.e. $f$ is of the form $ax+b$, where $a$ and $b$ are constants.
Notice that in the definitions above, we require the radius $r$ to run over all the positive numbers. Out of curiosity, I tried to find non harmonic functions which satisfy the MVPs only for $r=1$. It turns out any $1$-periodic function will satisfy the spherical MVP with $r=1$, which is obvious. But it seems much harder to find one for ball MVP. So here is my question:
Question: Does there exist a function $f$ which is not of the form $ax+b$, such that $\forall x\in\mathbb{R},$ $$\frac{1}{2}\int_{x-1}^{x+1}f(t)dt=f(x)?$$
Thank you!
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Just a side comment: for the mean value property to imply harmonicity, you don't need $r$ to run over all the positive numbers. It suffices (for example) that the property holds for all $r\in [0,\epsilon)$ for some $\epsilon > 0$. – Willie Wong♦ Mar 5 '12 at 9:34
@WillieWong: does $\epsilon$ depend on $x$? – Syang Chen Apr 13 '12 at 5:16
it may. For $C^2$ functions it follows directly from the proof for MVP to imply harmonicity. – Willie Wong♦ Apr 17 '12 at 4:30
@WillieWong: Can we weaken the $C^2$ assumption into, say $C^0$? In practice $C^2$ seems hard to verify. – Syang Chen Apr 28 '12 at 3:54
See page 17 of Harmonic Function Theory by Axler, Bourdon, and Ramey – Willie Wong♦ Apr 30 '12 at 8:29
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## 2 Answers
Note that
$$\frac{1}{2}\int_{x-1}^{x+1}e^{as}ds = \frac{\sinh(a)}{a}e^{ax}.$$
Let $a+ib \in \mathbb{C}$ be a non-zero root of $\sinh(z)=z$. Then also $a-ib$ is a root and because the MVP is linear the function $f(x) = e^{ax}\cos(bx)$ has the mean value property.
Remains to prove that $\sinh(z)-z$ has non-zero roots. I present two proofs below. The second one (that I gave first) is a bit clunky but has the advantage that it gives some information about the distribution of the roots.
Proof 1: Let $f(z) = \sinh(z)-z$. Then $f$ has an essential singularity at $\infty$. By Picard's great theorem the function $f$ attains all values infinitely often with at most one exception. In particular at least one of $0$ and $2\pi i$ must be attained infinitely often. However, $f(z + 2\pi i) = f(z) - 2\pi i$ and so both cases imply that $f$ has infinitely many roots.
Proof 2: I'll show that $|\sinh(z)|=|z|$ holds along some curve in $\mathbb{C}$ extending to $\infty$ and that $\sinh(z)/z$ must wind around the unit circle infinitely often along this curve.
If $z=x+iy$ then $2|\sinh(z)|^2 = \cosh(2x)-\cos(2y)$ and so $|\sinh(z)| = |z|$ exactly if $\cosh(2x)-2x^2=\cos(2y)+2y^2$. To show that for each $x$ there is a $y$ that satisfies this equation define the following functions:
$$f(x) = \cosh(2x)-2x^2, \ g(x) = \cos(2x)+2x^2.$$
On $\mathbb{R}_{>0}$ both functions are strictly increasing and $f(x) > g(x)$. (The latter inequality can be checked from their power series.) In particular, for each $x>0$ there exists a unique $y>x$ such that $f(x) = g(y)$. Let $\tau: \mathbb{R}_{>0} \rightarrow \mathbb{C}$ be the curve $\tau: x \mapsto x+iy$, then $|\tau'(x)| \geq 1$ and $|\sinh(\tau)| = |\tau|$. So the curve
$$x \mapsto \frac{\sinh(\tau)}{\tau}$$ maps into the unit circle. If we can bound its derivative from below, then it must wind around the unit circle (and therefore pass through $1$) infinitely many times. That this is indeed the case follows from the following inequalities:
$$\left|\frac{\partial}{\partial x}\frac{\sinh(\tau)}{\tau}\right| \geq \left| \frac{\cosh(\tau)}{\tau} - \frac{\sinh(\tau)}{\tau^2} \right| \geq \left| \frac{\sqrt{\left| |\tau|^2-1 \right|}}{|\tau|} - \frac{1}{|\tau|} \right| \xrightarrow{x \rightarrow \infty} 1$$
This shows that $\sinh(\tau) = \tau$ occurs infinitely many times. Moreover, all solutions of $\sinh(z) = z$ with $\Re(z)>0$ and $\Im(z) >0$ lie on $\tau$.
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Are you sure that $\sinh(z) = z$ has a nonzero solution? – Nick Strehlke Mar 4 '12 at 10:49
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@NickStrehlke Using Mathematica's function FindRootyou find many solutions like $z=13.9+3.35221\,I$ and $z=7.49768 + 2.76868$. I have not tried to prove that they are in fact solutions. – Julián Aguirre Mar 4 '12 at 11:54
@WimC: Nice! Could you show why $\sinh(z)=z$ has a nonzero solution? – Syang Chen Mar 4 '12 at 16:27
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@NickStrehlke Yes, I posted solutions to $\sin a=a$. For such $a$, $\sin(a x)$ and $\cos(a x)$ are a solutions to the problem. I was going to write the answer, but I WinC had already done so. – Julián Aguirre Mar 4 '12 at 20:05
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@WimC Even nicer proof for the existence of roots!! – Syang Chen Mar 4 '12 at 22:52
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We can construct many solutions by setting $f(x)=F'(x)$ where $F$ is smooth and satisfies $$2F'(x)=F(x+1)-F(x-1)$$ for all $x$. For example, $F$ can be generated from this formula by starting with an arbitrary smooth ($C^\infty$) function on $[-1,1]$ that vanishes near the points $-1$, $0$, and $1$: For $x>1$, $$F(x)= F(x-2)+2F'(x-1)$$ defines $F$ successively on $(k,k+1]$ for $k=1,2,\dots$, and for $x<-1$, $$F(x)=F(x+2)-2F'(x+1)$$ does the same trick successively on $[-k-1,-k)$. This yields a function $F$ that vanishes near each integer, and is clearly smooth.
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– Syang Chen Mar 5 '12 at 1:09
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http://mathoverflow.net/questions/108213/the-multiplicative-system-in-a-symmetric-monoidal-category
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## The multiplicative system in a symmetric monoidal category
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Let $\mathcal{C}$ be a symmetric monoidal category. In the 1973 paper "Note on monoidal localisation" by Brian Day, the multiplicative system of morphism in $\mathcal{C}$ has been discussed. See also this mathoverflow question by Martin Brandenburg.
My question is: can we consider a multiplicative system consists of both objects and morphisms in $\mathcal{C}$? This means that we have a collection of objects $x_i$ and a collection of morphisms $f_i$ such that $x_i \otimes x_j$ is still in the collection $x_i$ and $x_i$ and $f_j$ satisfies some "compatible condition". And can we define a localization along this more general multiplicative system?
Notice that in this viewpoint the case in the first paragraph can be considered as the multplicative system with only one object $1$ (and a system of morphisms).
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This is a bit above my categorical pay grade, so I will leave a hopefully helpful comment rather than an answer. A general strategy to working with objects in any category is to encode them via their identity morphisms. Is it enough in your case to use the theory of monoidal localization but with some identity morphisms in the mix? – Theo Johnson-Freyd Sep 27 at 14:02
@Theo: Yes I need some morphism in the mix. But still I'm interested in the case your mentioned: we consider a multiplicative system of objects and the identity morphisms of each object. Then what should be the requirement on the collection of objects to make them a multiplicative system? – Zhaoting Wei Sep 27 at 15:31
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http://mathoverflow.net/questions/17105/reference-for-the-expected-number-of-prime-factors-of-n-larger-than-nalpha-is-l/17114
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## Reference for the expected number of prime factors of n larger than n^alpha is -log alpha
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Let $0 < \alpha < 1$ be a constant. The expected number of prime factors of a "random" integer near $n$ which are greater than $n^\alpha$ is $-\log \alpha$.
It's my understanding that (properly formulated) this is a well-known fact in analytic number theory but I cannot find a reference for it. Can anybody provide a reference?
Edited to add (March 28):: The asymptotic density of positive integers $n$ with $k$th largest factor smaller than $n^{1/\alpha}$ is $\rho_k(\alpha)$, where we have $L_0(\alpha) = [\alpha > 0]$ and $$L_k(\alpha) = [\alpha \ge k] \int_k^\alpha L_{k-1}(t-1) \: {dt \over t},$$ and $1-\rho_k(\alpha) = \sum_{n=0}^\infty {-k \choose n} L_{n+k}(\alpha)$. (See Riesel, p. 162.) The density of positive integers with $k$th largest factor larger than $n^{1/\alpha}$ is therefore $1-\rho_k(\alpha)$, and so the expected number of factors larger than $n^{1/\alpha}$ is $\sum_{k \ge 1} (1-\rho_k(\alpha))$. Therefore the expected number of such factors is $$\sum_{k \ge 1} \sum_{n \ge 0} {-k \choose n} L_{n+k}(\alpha).$$ Letting $n+k = j$ we can rewrite this sum as $$\sum_{j \ge 1} \sum_{n=0}^{j-1} {n-j \choose n} L_j = \sum_{j \ge 1} L_j \left( \sum_{n=-0}^{j-1} (-1)^n {j-1 \choose n} \right)$$ and the inner sum is $0$ except when $j=1$, when it is $1$. So the expected number of factors larger than $n^{1/\alpha}$ is $L_1(\alpha)$; this is $\log \alpha$.
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## 4 Answers
Theorem 5.4 of Riesel, Prime Numbers and Computer Methods for Factorization, says "the number of prime factors $p$ of integers in the interval $[N-x,N+x]$ such that $a<\log\log p< b$ is proportional to $b-a$ if $b-a$ as well as $x$ are sufficiently large as $N\to\infty$."
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I am accepting this even though it is not exactly what I was looking for, because what I was looking for is a couple pages later in Riesel's book. – Michael Lugo Mar 28 2010 at 16:35
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You seem to be over complicating things. For $\alpha$ fixed, this is relatively easy to prove. I leave it to you deduce the desired result as a consequence of the following theorem:
Theorem: Let $x>0$ and fix $0<\alpha<1$. Define $\omega_{x,\alpha}(n)=\sum_{p|n,\ p>x^\alpha} 1$. Then this function has average value $$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=-\log \alpha +O\left(\frac{1}{\log x}\right).$$
Proof: Notice that
`$$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left[\frac{x}{p}\right] =\sum_{x^{\alpha}<p<\leq x}\frac{1}{p}-\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left\{\frac{x}{p}\right\}$$`
`$$= \sum_{x^\alpha<p<x} \frac{1}{p} +O\left(\frac{1}{\log x}\right)=\log \log (x)-\log \log x^\alpha +O\left(\frac{1}{\log x}\right)$$` `$$=-\log \alpha +O\left(\frac{1}{\log x}\right).$$`
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I believe you can extract this from a paper of Andrew Granville, "Prime divisors are Poisson distributed". There is an electronic copy of this on his website.
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Maybe. I feel like I've seen it stated explicitly, though. – Michael Lugo Mar 4 2010 at 18:01
Do you mean that $\log(1/\alpha)$ is the expected number of prime factors in $(x^\alpha,x]$ when $\alpha\to0$? For fixed $\alpha\in(0,1)$ what you are claiming is not true. For example,
$|{n\le x:\exists p|n\;{\rm with}\;\sqrt{x}\lt p\le x}|\sim x\log2$
and
$|{n\le x:p\le\sqrt{x}\;{\rm for all}\;p|n}|\sim(1-\log2)x$.
When $\alpha\to0$ it is not hard to prove what you need, but I am not sure where you can find a precise reference. For example, setting $\omega(n;y,z)=|{p|n:y\lt p\le z}|$ and following the proof of Theorem 6 in page 311 in Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory" gives that
$|{n\le x:|\omega(n;x^\alpha,x)-\log(1/\alpha)|\ge(1+\delta)\log(1/\alpha)}|\ll x\alpha^{Q(1+\delta)},$
where $Q(1+\delta)=\int_1^{1+\delta}\log tdt$.
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http://mathoverflow.net/questions/91454/simple-closed-curves-and-the-coefficent-of-expi-theta-in-the-associated-four/91784
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## Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series
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Given a continuous map $f:S^1\to \mathbb{C}$ from the unit circle to the complex numbers, one can form its Fourier series $\sum_{n=-\infty}^\infty a_n\exp(in\theta)$. I want to stick with those $f$ that give simple closed curves, bounding a closed topological disk, going round the disk in a counter-clockwise direction, and parametrized proportional to arclength. I am happy to add the hypothesis that $f'(t)$ is a continuous function of $t$ and that, for $t\in S^1$, $|f'(t)| = 1$.
Is it then true that $a_1\neq 0$?
If this is true, is $|a_1|$ bounded away from zero as $f$ varies? It may be that some other normalization might make the second question more tractable: for example, instead of normalizing the length to be $2\pi$ by a change of scale, as I have done above, one could require that a disk of unit radius be contained in the disk bounded by $f$. Any such normalization of $f$ would be highly acceptable.
I'm motivated by trying to describe the space of shapes'' in the plane, by using Fourier descriptors, a topic of interest both in machine vision and in microscopy in biology.
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Is $a_1 \neq 0$ the final goal or are you interested in necessary (and possible sufficient) conditions on the sequence $\{a_n\}$ for $$f(\theta) := \sum a_n e^{in\theta}$$ to be a Jordan curve? – alvarezpaiva Mar 19 2012 at 10:28
## 3 Answers
OK, let's modify Sean's construction to remove any doubts (it won't look the same, but it is based on the same idea). We will consider the curves symmetric with respect to the real axis and parametrized so that $f(-\theta)=\bar f(\theta)$, so we are sure that all Fourier coefficients are real. Now take $a\in\mathbb R$ and draw any continuous family of nice symmetric counterclockwise shapes $\Gamma_a$ that visit the points $1$, $a+i$, $a-i$ in this order. Note that the shapes will be necessarily non-convex for $a\ge 1$. Take small neighborhoods of these three points and replace the quick almost straight passages that are there by some "drunken walks" without self-intersections that have huge lengths but move essentially nowhere so that the whole length of the curve becomes essentially concentrated at those 3 points and the corresponding "wasted time" intervals are close to $(-\pi/2,\pi/2)$, $(\pi/2,\pi)$, $(-\pi,-\pi/2)$. Now, $2\pi a_1$ for the corresponding function is essentially $\int_{-\pi/2}^{\pi/2}\cos\theta\, d\theta+2\Re\left[(a+i)\int_{\pi/2}^{\pi}e^{-i\theta}\,d\theta\right]$, which is positive for large negative $a$ and negative for large positive $a$. However, the family of curves we created is continuous and so is the family of their parametrizations, so the intermediate value theorem finishes the story.
As usual, the existence of a counterexample most likely merely means that what you asked for is not what you need. So, what's the actual goal?
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Dear David,
This is just a reflection on your question:
Since you assume that the curve is parametrized by arc-length, applying Plancharel's formula to $f'$ yields $$\sum n^2|a_n|^2 = 1.$$ Moreover, you also assume that the map $f' : S^1 \rightarrow S^1$ has degree 1 and Brezis's formula for the degree of a $C^1$ map from the circle to the circle (Google Kahane's paper Winding number and Fourier series for the formula and the amusing story behind it) yields $$\sum n^3|a_n|^2 = 1.$$
Averaging these two equations and assuming $a_1 = 0$ one gets that $$\sum_{|n|> 1} {n^2(1 + n) \over 2}|a_n|^2 = 1$$ but I don't see right now if this and could lead to a contradiction with $\sum n^2|a_n|^2 = 1$.
I don't know if this helps with your precise question, but I think Brezis's formula for the degree in terms of Fourier coefficients could come in useful.
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Actually Brezis formula holds for maps in the Sobolev space $H^{1/2}(S^1,S^1)$ although I assumed that $f'$ is $C^1$. – alvarezpaiva Mar 17 2012 at 12:50
The answer to your first question is 'No'.
Let $g$ be the function $S^1\to S^1$ which starts at $1$, moves anticlockwise to $-1$, then moves clockwise $1 + \sqrt{2}$ times as fast once round the circle back to $-1$, and then moves anticlockwise back to $1$ again. This function $g$ has degree $0$ and $\hat{g}(0)=0$. The function $f(\theta) = g(\theta) e^{i\theta}$, which moves at a constant speed, therefore has degree $1$ and $\hat{f}(1)=0$.
You may reasonably complain at this point that $f$ is not differentiable and certainly not simple, but $f$ can be deformed very slightly so that it bounds a topological disc and makes smooth turns.
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I am missing why $g(\theta) e^{i \theta}$ has constant speed. The function $g$ makes one total loop clockwise and one total loop anti-clockwise. It makes the first loop 3 times as fast as the second, so it must spend $1/4$ of its time traveling clockwise. So $g$ is of the form $e^{4 i \theta}$ on the clockwise portions, and $e^{-(4/3) i \theta}$ on the anti-clockwise portions. So $g$ sometimes has speed $4-1=3$ and sometime has speed $1-(-4/3) = 7/3$. – David Speyer Mar 17 2012 at 17:05
You're absolutely right. I guess what I want is the following: suppose that $g$ moves clockwise $x$ times as fast as it moves anticlockwise. Then it spends $1/(x+1)$ of its time clockwise and $x/(x+1)$ of its time anticlockwise. Its speed clockwise is therefore $1+x$, anticlockwise $1+1/x$. I want then that $1+x−1=1+1/x+1$, i.e., $x^2−2x−1=0$. – Sean Eberhard Mar 17 2012 at 17:41
Editted solution to reflect this. – Sean Eberhard Mar 17 2012 at 17:43
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Sean's example is beautifully simple. It definitely shows that $|a_1|$ cannot be bounded away from zero. However, deforming $f$ so that it becomes a simple closed curve, with $a_1=0$ exactly, seems delicate to me. Perhaps I'm not looking at things the right way. Because I'm still hoping for a definitive answer that satisfies all of my conditions, I'm not yet ready to mark Sean's as RIGHT, even though I'm full of admiration for its simplicity and brevity. – David Epstein Mar 18 2012 at 21:54
I agree with you that this is a little delicate. I suppose it is more of a belief that $f$ can be deformed appropriately while maintaining $a_1=0$. If I think of a simple argument I will relate it. – Sean Eberhard Mar 18 2012 at 22:16
show 1 more comment
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http://nrich.maths.org/6690/solution
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### Rotating Triangle
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Doodles
A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not 'triple points'. Number the vertex points in any order. Starting at any point on the doodle, trace it until you get back to where you started. Write down the numbers of the vertices as you pass through them. So you have a [not necessarily unique] list of numbers for each doodle. Prove that 1)each vertex number in a list occurs twice. [easy!] 2)between each pair of vertex numbers in a list there are an even number of other numbers [hard!]
### Russian Cubes
How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted in order 'blue then red then blue then red then blue then red'. Having finished one cube, they begin to paint the next one. Prove that the girl can choose the faces she paints so as to make the second cube the same as the first.
# Tower of Hanoi
##### Stage: 4 Challenge Level:
Matthew, from Verulam School, made the following observations:
Question B helps you work out the difference between the numbers: with 1 disc the moves done will be 1, with 2 discs the moves done will be 3, with 3 discs the moves done will be 7, and with 4 discs the moves done will be 15.
There's a pattern: the difference between the 1st and the 2nd disc is 2, the difference between the 2nd and 3rd disc is 4, and the difference between the 3rd and 4th disc is 8 - it's just doubling the difference each time.
Tom, from Wilson's School worked out a formula for the Final Challenge and had a go at the extension:
If there are $n$ discs on the tower to find out the number of moves you need to work out $2^{n-1}$ then multiply your answer by $2$ and then subtract $1$
$2(2^{n-1}) -1$
Extension:
$64$ discs would be $2^{64-1} = 2^{63} = 9,223,372,036,854,775,808$
Multiply by $2: 18,446,744,073,709,551,616$,
Subtract $1: 18,446,744,073,709,551,615$ seconds
Divide by $60: 307,445,734,561,825,860.25$ minutes
Divide by $60: 5,124,095,576,030,431.004$ hours (3 d.p.)
Divide by $24: 213,503,982,334,601.292$ days (3 d.p.)
Divide by $365.25: 584,542,046,090.626$ years (3 d.p.)
It will take roughly 584.5 billion years from the start of time.
As the universe is approximately 13.7 billion years old now I don't think we need to worry yet!
Miltoon explained how to generate the formula:
From watching the video, I see that for two discs, $3$ moves are needed. For three discs, I first see $2$ discs being removed, and re-piled, therefore, three steps are used. Now, the largest disc is moved to the pole at the end. Now the two discs have to be re-piled again, on top of that largest disc, which also takes $3$ moves. So in total, $3 + 3 + 1$ moves are needed, because we re-piled the 'two discs' twice ($6$ moves), and moved the largest disc once, leaving us with $7$ moves.
If you have $D$ discs, and you know the amount of moves needed for $D$ discs - let's call it $F$, then if you want to calculate the number of moves needed for $D + 1$ discs, physically your tower will have a new base - and you move your $D$ discs out first, and re-pile them, which takes $F$ moves, you move the new, large disc, and put it at the furthest pole, which increased the number of moves one unit, and then you re-pile the $D$ disc tower again, but this time on top of the new large disc, so in total, it takes $2F+1$ moves for a tower with $D+1$ discs.
Now I can be sure that the sequence goes: $1, (2\times1 + 1) = 3, (2\times3 + 1) = 7, (2\times7 + 1) = 15 \dots$
Therefore, the formula is just $2^D - 1$ where $D$ is the number of discs.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness/58544
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## formal smoothness versus reducedness
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi,
I have the following situation: $R,H$ schemes (can be assumed noetherian and of finite type) over a field $k$ which we can assume to be algebraically closed, with $H$ reduced, $Y\subset R\times \mathbb{P}_k^n$ an open subset, $p:Y\rightarrow R$ the restriction of the projection onto the first factor and $w:Y\rightarrow H$ a surjective formally smooth morphism. How can I show that $R$ is reduced? Thank you
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Dear unknown, if R is the disjoint union of a point and a point with multiplicity two, H is a point, n=0, Y is the reduced point of (R cross P^0), and w an isomorphism, then your situation seems to hold but R is not reduced. Perhaps I am missing something, eg R connected? David – David Holmes Mar 15 2011 at 14:39
@David Holmes why $Y$ smooth$/ H$ should imply $R$ reduced? – unknown Mar 15 2011 at 14:44
1
Are you missing something in your question? You don't seem to use the morphism $p$ at all. As it stands it appears false: if $R$ is, say, a curve with an embedded point $P$, then take $Y$ to be the complement of $p^{-1}(P)$ in $R \times \mathbb{P}^1$, take $H=Y$ and $w$ the identity morphism. – Martin Bright Mar 15 2011 at 14:44
@David Holmes yes, can assume $R$ connected – unknown Mar 15 2011 at 14:46
@Martin Bright ok, what about if we assume also $p$ formally smooth or smooth? – unknown Mar 15 2011 at 14:51
show 4 more comments
## 1 Answer
Since $H$ is reduced and $Y$ is smooth over $H$ (I am assuming that everything is finite type over $k$, so smooth and formally smooth are the same) we see that $Y$ is reduced.
So the problem is the following: show that if $Y \subset R \times \mathbb P^n$ is open and reduced, and the projection $Y \to R$ is surjective (taking into account the remark to this effect in the comments), then $R$ is reduced.
Here is the proof: Let $x$ be a point of $R$, and let $y$ be a point of $Y$ lying over $x$. Recalling that $\mathbb P^n$ is the union of $n + 1$ open subsets isomorphic to $\mathbb A^n$, we may assume that $y \in R\times \mathbb A^n$ (for an appropriate choice of one of these $n+1$ copies). The stalk $\mathcal O_{Y,y}$ is then equal to a localization of $\mathcal O_{R,x}[x_1,\ldots,x_n]$. It is reduced by assumption, and so $\mathcal O_{R,x}$ is reduced. Since $x \in R$ was arbitrary, we see that $R$ is reduced.
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